二叉树
目录：
8、【剑指Offer学习】【面试题08：二叉树的下一个结点】
32、【剑指Offer学习】【面试题32：不分行从上往下打印二叉树】
33、【剑指Offer学习】【面试题33：二叉搜索树的后序遍历序列】
34、【剑指Offer学习】【面试题34：二叉树中和为某一值的路径】
36、【剑指Offer学习】【面试题36：二叉搜索树与双向链表】
54、【剑指Offer学习】【面试题54：二叉搜索树的第k个结点】
68、【剑指Offer学习】【面试题68：树中两个结点的最低公共祖先】
7. 【剑指Offer学习】【面试题07：重建二叉树】
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题7：重建二叉树
// 题目：输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输
// 入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,
// 2, 4, 7, 3, 5, 6, 8}和中序遍历序列{4, 7, 2, 1, 5, 3, 8, 6}，则重建出
// 图2.6所示的二叉树并输出它的头结点。
#include "..\Utilities\BinaryTree.h"
#include <exception>
#include <cstdio>
BinaryTreeNode* ConstructCore(int* startPreorder, int* endPreorder, int* startInorder, int* endInorder);
BinaryTreeNode* Construct(int* preorder, int* inorder, int length)
{
if(preorder == nullptr || inorder == nullptr || length <= 0)
return nullptr;
return ConstructCore(preorder, preorder + length - 1,
inorder, inorder + length - 1);
}
BinaryTreeNode* ConstructCore
(
int* startPreorder, int* endPreorder,
int* startInorder, int* endInorder
)
{
// 前序遍历序列的第一个数字是根结点的值
int rootValue = startPreorder[0];
BinaryTreeNode* root = new BinaryTreeNode();
root->m_nValue = rootValue;
root->m_pLeft = root->m_pRight = nullptr;
if(startPreorder == endPreorder)
{
if(startInorder == endInorder && *startPreorder == *startInorder)
return root;
else
throw std::exception("Invalid input.");
}
// 在中序遍历中找到根结点的值
int* rootInorder = startInorder;
while(rootInorder <= endInorder && *rootInorder != rootValue)
++ rootInorder;
if(rootInorder == endInorder && *rootInorder != rootValue)
throw std::exception("Invalid input.");
int leftLength = rootInorder - startInorder;
int* leftPreorderEnd = startPreorder + leftLength;
if(leftLength > 0)
{
// 构建左子树
root->m_pLeft = ConstructCore(startPreorder + 1, leftPreorderEnd,
startInorder, rootInorder - 1);
}
if(leftLength < endPreorder - startPreorder)
{
// 构建右子树
root->m_pRight = ConstructCore(leftPreorderEnd + 1, endPreorder,
rootInorder + 1, endInorder);
}
return root;
}
// ====================测试代码====================
void Test(char* testName, int* preorder, int* inorder, int length)
{
if(testName != nullptr)
printf("%s begins:\n", testName);
printf("The preorder sequence is: ");
for(int i = 0; i < length; ++ i)
printf("%d ", preorder[i]);
printf("\n");
printf("The inorder sequence is: ");
for(int i = 0; i < length; ++ i)
printf("%d ", inorder[i]);
printf("\n");
try
{
BinaryTreeNode* root = Construct(preorder, inorder, length);
PrintTree(root);
DestroyTree(root);
}
catch(std::exception& exception)
{
printf("Invalid Input.\n");
}
}
// 普通二叉树
// 1
// / \
// 2 3
// / / \
// 4 5 6
// \ /
// 7 8
void Test1()
{
const int length = 8;
int preorder[length] = {1, 2, 4, 7, 3, 5, 6, 8};
int inorder[length] = {4, 7, 2, 1, 5, 3, 8, 6};
Test("Test1", preorder, inorder, length);
}
// 所有结点都没有右子结点
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
void Test2()
{
const int length = 5;
int preorder[length] = {1, 2, 3, 4, 5};
int inorder[length] = {5, 4, 3, 2, 1};
Test("Test2", preorder, inorder, length);
}
// 所有结点都没有左子结点
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void Test3()
{
const int length = 5;
int preorder[length] = {1, 2, 3, 4, 5};
int inorder[length] = {1, 2, 3, 4, 5};
Test("Test3", preorder, inorder, length);
}
// 树中只有一个结点
void Test4()
{
const int length = 1;
int preorder[length] = {1};
int inorder[length] = {1};
Test("Test4", preorder, inorder, length);
}
// 完全二叉树
// 1
// / \
// 2 3
// / \ / \
// 4 5 6 7
void Test5()
{
const int length = 7;
int preorder[length] = {1, 2, 4, 5, 3, 6, 7};
int inorder[length] = {4, 2, 5, 1, 6, 3, 7};
Test("Test5", preorder, inorder, length);
}
// 输入空指针
void Test6()
{
Test("Test6", nullptr, nullptr, 0);
}
// 输入的两个序列不匹配
void Test7()
{
const int length = 7;
int preorder[length] = {1, 2, 4, 5, 3, 6, 7};
int inorder[length] = {4, 2, 8, 1, 6, 3, 7};
Test("Test7: for unmatched input", preorder, inorder, length);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
return 0;
}
8. 【剑指Offer学习】【面试题08：二叉树的下一个结点】
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题8：二叉树的下一个结点
// 题目：给定一棵二叉树和其中的一个结点，如何找出中序遍历顺序的下一个结点？
// 树中的结点除了有两个分别指向左右子结点的指针以外，还有一个指向父结点的指针。
#include <stdio.h>
struct BinaryTreeNode
{
int m_nValue;
BinaryTreeNode* m_pLeft;
BinaryTreeNode* m_pRight;
BinaryTreeNode* m_pParent;
};
BinaryTreeNode* GetNext(BinaryTreeNode* pNode)
{
if(pNode == nullptr)
return nullptr;
BinaryTreeNode* pNext = nullptr;
if(pNode->m_pRight != nullptr)
{
BinaryTreeNode* pRight = pNode->m_pRight;
while(pRight->m_pLeft != nullptr)
pRight = pRight->m_pLeft;
pNext = pRight;
}
else if(pNode->m_pParent != nullptr)
{
BinaryTreeNode* pCurrent = pNode;
BinaryTreeNode* pParent = pNode->m_pParent;
while(pParent != nullptr && pCurrent == pParent->m_pRight)
{
pCurrent = pParent;
pParent = pParent->m_pParent;
}
pNext = pParent;
}
return pNext;
}
// ==================== 辅助代码用来构建二叉树 ====================
BinaryTreeNode* CreateBinaryTreeNode(int value)
{
BinaryTreeNode* pNode = new BinaryTreeNode();
pNode->m_nValue = value;
pNode->m_pLeft = nullptr;
pNode->m_pRight = nullptr;
pNode->m_pParent = nullptr;
return pNode;
}
void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
if(pParent != nullptr)
{
pParent->m_pLeft = pLeft;
pParent->m_pRight = pRight;
if(pLeft != nullptr)
pLeft->m_pParent = pParent;
if(pRight != nullptr)
pRight->m_pParent = pParent;
}
}
void PrintTreeNode(BinaryTreeNode* pNode)
{
if(pNode != nullptr)
{
printf("value of this node is: %d\n", pNode->m_nValue);
if(pNode->m_pLeft != nullptr)
printf("value of its left child is: %d.\n", pNode->m_pLeft->m_nValue);
else
printf("left child is null.\n");
if(pNode->m_pRight != nullptr)
printf("value of its right child is: %d.\n", pNode->m_pRight->m_nValue);
else
printf("right child is null.\n");
}
else
{
printf("this node is null.\n");
}
printf("\n");
}
void PrintTree(BinaryTreeNode* pRoot)
{
PrintTreeNode(pRoot);
if(pRoot != nullptr)
{
if(pRoot->m_pLeft != nullptr)
PrintTree(pRoot->m_pLeft);
if(pRoot->m_pRight != nullptr)
PrintTree(pRoot->m_pRight);
}
}
void DestroyTree(BinaryTreeNode* pRoot)
{
if(pRoot != nullptr)
{
BinaryTreeNode* pLeft = pRoot->m_pLeft;
BinaryTreeNode* pRight = pRoot->m_pRight;
delete pRoot;
pRoot = nullptr;
DestroyTree(pLeft);
DestroyTree(pRight);
}
}
// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pNode, BinaryTreeNode* expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
BinaryTreeNode* pNext = GetNext(pNode);
if(pNext == expected)
printf("Passed.\n");
else
printf("FAILED.\n");
}
// 8
// 6 10
// 5 7 9 11
void Test1_7()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);
ConnectTreeNodes(pNode8, pNode6, pNode10);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);
Test("Test1", pNode8, pNode9);
Test("Test2", pNode6, pNode7);
Test("Test3", pNode10, pNode11);
Test("Test4", pNode5, pNode6);
Test("Test5", pNode7, pNode8);
Test("Test6", pNode9, pNode10);
Test("Test7", pNode11, nullptr);
DestroyTree(pNode8);
}
// 5
// 4
// 3
// 2
void Test8_11()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
Test("Test8", pNode5, nullptr);
Test("Test9", pNode4, pNode5);
Test("Test10", pNode3, pNode4);
Test("Test11", pNode2, pNode3);
DestroyTree(pNode5);
}
// 2
// 3
// 4
// 5
void Test12_15()
{
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
Test("Test12", pNode5, nullptr);
Test("Test13", pNode4, pNode5);
Test("Test14", pNode3, pNode4);
Test("Test15", pNode2, pNode3);
DestroyTree(pNode2);
}
void Test16()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
Test("Test16", pNode5, nullptr);
DestroyTree(pNode5);
}
int main(int argc, char* argv[])
{
Test1_7();
Test8_11();
Test12_15();
Test16();
}
26. 面试题26：树的子结构
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题26：树的子结构
// 题目：输入两棵二叉树A和B，判断B是不是A的子结构。
#include <cstdio>
struct BinaryTreeNode
{
double m_dbValue;
BinaryTreeNode* m_pLeft;
BinaryTreeNode* m_pRight;
};
bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2);
bool Equal(double num1, double num2);
bool HasSubtree(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
bool result = false;
if(pRoot1 != nullptr && pRoot2 != nullptr)
{
if(Equal(pRoot1->m_dbValue, pRoot2->m_dbValue))
result = DoesTree1HaveTree2(pRoot1, pRoot2);
if(!result)
result = HasSubtree(pRoot1->m_pLeft, pRoot2);
if(!result)
result = HasSubtree(pRoot1->m_pRight, pRoot2);
}
return result;
}
bool DoesTree1HaveTree2(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
if(pRoot2 == nullptr)
return true;
if(pRoot1 == nullptr)
return false;
if(!Equal(pRoot1->m_dbValue, pRoot2->m_dbValue))
return false;
return DoesTree1HaveTree2(pRoot1->m_pLeft, pRoot2->m_pLeft) &&
DoesTree1HaveTree2(pRoot1->m_pRight, pRoot2->m_pRight);
}
bool Equal(double num1, double num2)
{
if((num1 - num2 > -0.0000001) && (num1 - num2 < 0.0000001))
return true;
else
return false;
}
// ====================辅助测试代码====================
BinaryTreeNode* CreateBinaryTreeNode(double dbValue)
{
BinaryTreeNode* pNode = new BinaryTreeNode();
pNode->m_dbValue = dbValue;
pNode->m_pLeft = nullptr;
pNode->m_pRight = nullptr;
return pNode;
}
void ConnectTreeNodes(BinaryTreeNode* pParent, BinaryTreeNode* pLeft, BinaryTreeNode* pRight)
{
if(pParent != nullptr)
{
pParent->m_pLeft = pLeft;
pParent->m_pRight = pRight;
}
}
void DestroyTree(BinaryTreeNode* pRoot)
{
if(pRoot != nullptr)
{
BinaryTreeNode* pLeft = pRoot->m_pLeft;
BinaryTreeNode* pRight = pRoot->m_pRight;
delete pRoot;
pRoot = nullptr;
DestroyTree(pLeft);
DestroyTree(pRight);
}
}
// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2, bool expected)
{
if(HasSubtree(pRoot1, pRoot2) == expected)
printf("%s passed.\n", testName);
else
printf("%s failed.\n", testName);
}
// 树中结点含有分叉，树B是树A的子结构
// 8 8
// / \ / \
// 8 7 9 2
// / \
// 9 2
// / \
// 4 7
void Test1()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA6 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNodeA7 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);
BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3);
Test("Test1", pNodeA1, pNodeB1, true);
DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}
// 树中结点含有分叉，树B不是树A的子结构
// 8 8
// / \ / \
// 8 7 9 2
// / \
// 9 3
// / \
// 4 7
void Test2()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNodeA6 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNodeA7 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3);
ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5);
ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7);
BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3);
Test("Test2", pNodeA1, pNodeB1, false);
DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}
// 树中结点只有左子结点，树B是树A的子结构
// 8 8
// / /
// 8 9
// / /
// 9 2
// /
// 2
// /
// 5
void Test3()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNodeA1, pNodeA2, nullptr);
ConnectTreeNodes(pNodeA2, pNodeA3, nullptr);
ConnectTreeNodes(pNodeA3, pNodeA4, nullptr);
ConnectTreeNodes(pNodeA4, pNodeA5, nullptr);
BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNodeB1, pNodeB2, nullptr);
ConnectTreeNodes(pNodeB2, pNodeB3, nullptr);
Test("Test3", pNodeA1, pNodeB1, true);
DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}
// 树中结点只有左子结点，树B不是树A的子结构
// 8 8
// / /
// 8 9
// / /
// 9 3
// /
// 2
// /
// 5
void Test4()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNodeA1, pNodeA2, nullptr);
ConnectTreeNodes(pNodeA2, pNodeA3, nullptr);
ConnectTreeNodes(pNodeA3, pNodeA4, nullptr);
ConnectTreeNodes(pNodeA4, pNodeA5, nullptr);
BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(3);
ConnectTreeNodes(pNodeB1, pNodeB2, nullptr);
ConnectTreeNodes(pNodeB2, pNodeB3, nullptr);
Test("Test4", pNodeA1, pNodeB1, false);
DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}
// 树中结点只有右子结点，树B是树A的子结构
// 8 8
// \ \
// 8 9
// \ \
// 9 2
// \
// 2
// \
// 5
void Test5()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNodeA1, nullptr, pNodeA2);
ConnectTreeNodes(pNodeA2, nullptr, pNodeA3);
ConnectTreeNodes(pNodeA3, nullptr, pNodeA4);
ConnectTreeNodes(pNodeA4, nullptr, pNodeA5);
BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNodeB1, nullptr, pNodeB2);
ConnectTreeNodes(pNodeB2, nullptr, pNodeB3);
Test("Test5", pNodeA1, pNodeB1, true);
DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}
// 树A中结点只有右子结点，树B不是树A的子结构
// 8 8
// \ \
// 8 9
// \ / \
// 9 3 2
// \
// 2
// \
// 5
void Test6()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNodeA5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNodeA1, nullptr, pNodeA2);
ConnectTreeNodes(pNodeA2, nullptr, pNodeA3);
ConnectTreeNodes(pNodeA3, nullptr, pNodeA4);
ConnectTreeNodes(pNodeA4, nullptr, pNodeA5);
BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNodeB4 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNodeB1, nullptr, pNodeB2);
ConnectTreeNodes(pNodeB2, pNodeB3, pNodeB4);
Test("Test6", pNodeA1, pNodeB1, false);
DestroyTree(pNodeA1);
DestroyTree(pNodeB1);
}
// 树A为空树
void Test7()
{
BinaryTreeNode* pNodeB1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeB2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeB3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNodeB4 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNodeB1, nullptr, pNodeB2);
ConnectTreeNodes(pNodeB2, pNodeB3, pNodeB4);
Test("Test7", nullptr, pNodeB1, false);
DestroyTree(pNodeB1);
}
// 树B为空树
void Test8()
{
BinaryTreeNode* pNodeA1 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNodeA2 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNodeA3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNodeA4 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNodeA1, nullptr, pNodeA2);
ConnectTreeNodes(pNodeA2, pNodeA3, pNodeA4);
Test("Test8", pNodeA1, nullptr, false);
DestroyTree(pNodeA1);
}
// 树A和树B都为空
void Test9()
{
Test("Test9", nullptr, nullptr, false);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
Test8();
Test9();
return 0;
}
27. 面试题27：二叉树的镜像
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题27：二叉树的镜像
// 题目：请完成一个函数，输入一个二叉树，该函数输出它的镜像。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
#include <stack>
void MirrorRecursively(BinaryTreeNode *pNode)
{
if((pNode == nullptr) || (pNode->m_pLeft == nullptr && pNode->m_pRight))
return;
BinaryTreeNode *pTemp = pNode->m_pLeft;
pNode->m_pLeft = pNode->m_pRight;
pNode->m_pRight = pTemp;
if(pNode->m_pLeft)
MirrorRecursively(pNode->m_pLeft);
if(pNode->m_pRight)
MirrorRecursively(pNode->m_pRight);
}
void MirrorIteratively(BinaryTreeNode* pRoot)
{
if(pRoot == nullptr)
return;
std::stack<BinaryTreeNode*> stackTreeNode;
stackTreeNode.push(pRoot);
while(stackTreeNode.size() > 0)
{
BinaryTreeNode *pNode = stackTreeNode.top();
stackTreeNode.pop();
BinaryTreeNode *pTemp = pNode->m_pLeft;
pNode->m_pLeft = pNode->m_pRight;
pNode->m_pRight = pTemp;
if(pNode->m_pLeft)
stackTreeNode.push(pNode->m_pLeft);
if(pNode->m_pRight)
stackTreeNode.push(pNode->m_pRight);
}
}
// ====================测试代码====================
// 测试完全二叉树：除了叶子节点，其他节点都有两个子节点
// 8
// 6 10
// 5 7 9 11
void Test1()
{
printf("=====Test1 starts:=====\n");
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);
ConnectTreeNodes(pNode8, pNode6, pNode10);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);
PrintTree(pNode8);
printf("=====Test1: MirrorRecursively=====\n");
MirrorRecursively(pNode8);
PrintTree(pNode8);
printf("=====Test1: MirrorIteratively=====\n");
MirrorIteratively(pNode8);
PrintTree(pNode8);
DestroyTree(pNode8);
}
// 测试二叉树：出叶子结点之外，左右的结点都有且只有一个左子结点
// 8
// 7
// 6
// 5
// 4
void Test2()
{
printf("=====Test2 starts:=====\n");
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
ConnectTreeNodes(pNode8, pNode7, nullptr);
ConnectTreeNodes(pNode7, pNode6, nullptr);
ConnectTreeNodes(pNode6, pNode5, nullptr);
ConnectTreeNodes(pNode5, pNode4, nullptr);
PrintTree(pNode8);
printf("=====Test2: MirrorRecursively=====\n");
MirrorRecursively(pNode8);
PrintTree(pNode8);
printf("=====Test2: MirrorIteratively=====\n");
MirrorIteratively(pNode8);
PrintTree(pNode8);
DestroyTree(pNode8);
}
// 测试二叉树：出叶子结点之外，左右的结点都有且只有一个右子结点
// 8
// 7
// 6
// 5
// 4
void Test3()
{
printf("=====Test3 starts:=====\n");
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
ConnectTreeNodes(pNode8, nullptr, pNode7);
ConnectTreeNodes(pNode7, nullptr, pNode6);
ConnectTreeNodes(pNode6, nullptr, pNode5);
ConnectTreeNodes(pNode5, nullptr, pNode4);
PrintTree(pNode8);
printf("=====Test3: MirrorRecursively=====\n");
MirrorRecursively(pNode8);
PrintTree(pNode8);
printf("=====Test3: MirrorIteratively=====\n");
MirrorIteratively(pNode8);
PrintTree(pNode8);
DestroyTree(pNode8);
}
// 测试空二叉树：根结点为空指针
void Test4()
{
printf("=====Test4 starts:=====\n");
BinaryTreeNode* pNode = nullptr;
PrintTree(pNode);
printf("=====Test4: MirrorRecursively=====\n");
MirrorRecursively(pNode);
PrintTree(pNode);
printf("=====Test4: MirrorIteratively=====\n");
MirrorIteratively(pNode);
PrintTree(pNode);
}
// 测试只有一个结点的二叉树
void Test5()
{
printf("=====Test5 starts:=====\n");
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
PrintTree(pNode8);
printf("=====Test4: MirrorRecursively=====\n");
MirrorRecursively(pNode8);
PrintTree(pNode8);
printf("=====Test4: MirrorIteratively=====\n");
MirrorIteratively(pNode8);
PrintTree(pNode8);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
return 0;
}
28. 面试题28：对称的二叉树
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题28：对称的二叉树
// 题目：请实现一个函数，用来判断一棵二叉树是不是对称的。如果一棵二叉树和
// 它的镜像一样，那么它是对称的。
#include <cstdio>
#include "../Utilities/BinaryTree.h"
bool isSymmetrical(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2);
bool isSymmetrical(BinaryTreeNode* pRoot)
{
return isSymmetrical(pRoot, pRoot);
}
bool isSymmetrical(BinaryTreeNode* pRoot1, BinaryTreeNode* pRoot2)
{
if(pRoot1 == nullptr && pRoot2 == nullptr)
return true;
if(pRoot1 == nullptr || pRoot2 == nullptr)
return false;
if(pRoot1->m_nValue != pRoot2->m_nValue)
return false;
return isSymmetrical(pRoot1->m_pLeft, pRoot2->m_pRight)
&& isSymmetrical(pRoot1->m_pRight, pRoot2->m_pLeft);
}
// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pRoot, bool expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
if(isSymmetrical(pRoot) == expected)
printf("Passed.\n");
else
printf("FAILED.\n");
}
// 8
// 6 6
// 5 7 7 5
void Test1()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode61 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode62 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode51 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode71 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode72 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode52 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode8, pNode61, pNode62);
ConnectTreeNodes(pNode61, pNode51, pNode71);
ConnectTreeNodes(pNode62, pNode72, pNode52);
Test("Test1", pNode8, true);
DestroyTree(pNode8);
}
// 8
// 6 9
// 5 7 7 5
void Test2()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode61 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode51 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode71 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode72 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode52 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode8, pNode61, pNode9);
ConnectTreeNodes(pNode61, pNode51, pNode71);
ConnectTreeNodes(pNode9, pNode72, pNode52);
Test("Test2", pNode8, false);
DestroyTree(pNode8);
}
// 8
// 6 6
// 5 7 7
void Test3()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode61 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode62 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode51 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode71 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode72 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNode8, pNode61, pNode62);
ConnectTreeNodes(pNode61, pNode51, pNode71);
ConnectTreeNodes(pNode62, pNode72, nullptr);
Test("Test3", pNode8, false);
DestroyTree(pNode8);
}
// 5
// / \
// 3 3
// / \
// 4 4
// / \
// 2 2
// / \
// 1 1
void Test4()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode31 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode32 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode41 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode42 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode21 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode22 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode12 = CreateBinaryTreeNode(1);
ConnectTreeNodes(pNode5, pNode31, pNode32);
ConnectTreeNodes(pNode31, pNode41, nullptr);
ConnectTreeNodes(pNode32, nullptr, pNode42);
ConnectTreeNodes(pNode41, pNode21, nullptr);
ConnectTreeNodes(pNode42, nullptr, pNode22);
ConnectTreeNodes(pNode21, pNode11, nullptr);
ConnectTreeNodes(pNode22, nullptr, pNode12);
Test("Test4", pNode5, true);
DestroyTree(pNode5);
}
// 5
// / \
// 3 3
// / \
// 4 4
// / \
// 6 2
// / \
// 1 1
void Test5()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode31 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode32 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode41 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode42 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode22 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode12 = CreateBinaryTreeNode(1);
ConnectTreeNodes(pNode5, pNode31, pNode32);
ConnectTreeNodes(pNode31, pNode41, nullptr);
ConnectTreeNodes(pNode32, nullptr, pNode42);
ConnectTreeNodes(pNode41, pNode6, nullptr);
ConnectTreeNodes(pNode42, nullptr, pNode22);
ConnectTreeNodes(pNode6, pNode11, nullptr);
ConnectTreeNodes(pNode22, nullptr, pNode12);
Test("Test5", pNode5, false);
DestroyTree(pNode5);
}
// 5
// / \
// 3 3
// / \
// 4 4
// / \
// 2 2
// \
// 1
void Test6()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode31 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode32 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode41 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode42 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode21 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode22 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode12 = CreateBinaryTreeNode(1);
ConnectTreeNodes(pNode5, pNode31, pNode32);
ConnectTreeNodes(pNode31, pNode41, nullptr);
ConnectTreeNodes(pNode32, nullptr, pNode42);
ConnectTreeNodes(pNode41, pNode21, nullptr);
ConnectTreeNodes(pNode42, nullptr, pNode22);
ConnectTreeNodes(pNode21, nullptr, nullptr);
ConnectTreeNodes(pNode22, nullptr, pNode12);
Test("Test6", pNode5, false);
DestroyTree(pNode5);
}
// 只有一个结点
void Test7()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
Test("Test7", pNode1, true);
DestroyTree(pNode1);
}
// 没有结点
void Test8()
{
Test("Test8", nullptr, true);
}
// 所有结点都有相同的值，树对称
// 5
// / \
// 5 5
// / \
// 5 5
// / \
// 5 5
void Test9()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode21 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode22 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode31 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode32 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode41 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode42 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, pNode21, pNode22);
ConnectTreeNodes(pNode21, pNode31, nullptr);
ConnectTreeNodes(pNode22, nullptr, pNode32);
ConnectTreeNodes(pNode31, pNode41, nullptr);
ConnectTreeNodes(pNode32, nullptr, pNode42);
ConnectTreeNodes(pNode41, nullptr, nullptr);
ConnectTreeNodes(pNode42, nullptr, nullptr);
Test("Test9", pNode1, true);
DestroyTree(pNode1);
}
// 所有结点都有相同的值，树不对称
// 5
// / \
// 5 5
// / \
// 5 5
// / /
// 5 5
void Test10()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode21 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode22 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode31 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode32 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode41 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode42 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, pNode21, pNode22);
ConnectTreeNodes(pNode21, pNode31, nullptr);
ConnectTreeNodes(pNode22, nullptr, pNode32);
ConnectTreeNodes(pNode31, pNode41, nullptr);
ConnectTreeNodes(pNode32, pNode42, nullptr);
ConnectTreeNodes(pNode41, nullptr, nullptr);
ConnectTreeNodes(pNode42, nullptr, nullptr);
Test("Test10", pNode1, false);
DestroyTree(pNode1);
}
void main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
Test8();
Test9();
Test10();
}
}
32. 面试题32：不分行从上往下打印二叉树
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题32（一）：不分行从上往下打印二叉树
// 题目：从上往下打印出二叉树的每个结点，同一层的结点按照从左到右的顺序打印。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
#include <deque>
void PrintFromTopToBottom(BinaryTreeNode* pRoot)
{
if(pRoot == nullptr)
return;
std::deque<BinaryTreeNode *> dequeTreeNode;
dequeTreeNode.push_back(pRoot);
while(dequeTreeNode.size())
{
BinaryTreeNode *pNode = dequeTreeNode.front();
dequeTreeNode.pop_front();
printf("%d ", pNode->m_nValue);
if(pNode->m_pLeft)
dequeTreeNode.push_back(pNode->m_pLeft);
if(pNode->m_pRight)
dequeTreeNode.push_back(pNode->m_pRight);
}
}
// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pRoot)
{
if(testName != nullptr)
printf("%s begins: \n", testName);
PrintTree(pRoot);
printf("The nodes from top to bottom, from left to right are: \n");
PrintFromTopToBottom(pRoot);
printf("\n\n");
}
// 10
// / \
// 6 14
// /\ /\
// 4 8 12 16
void Test1()
{
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode14 = CreateBinaryTreeNode(14);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode12 = CreateBinaryTreeNode(12);
BinaryTreeNode* pNode16 = CreateBinaryTreeNode(16);
ConnectTreeNodes(pNode10, pNode6, pNode14);
ConnectTreeNodes(pNode6, pNode4, pNode8);
ConnectTreeNodes(pNode14, pNode12, pNode16);
Test("Test1", pNode10);
DestroyTree(pNode10);
}
// 5
// /
// 4
// /
// 3
// /
// 2
// /
// 1
void Test2()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
ConnectTreeNodes(pNode2, pNode1, nullptr);
Test("Test2", pNode5);
DestroyTree(pNode5);
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void Test3()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, nullptr, pNode2);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
Test("Test3", pNode1);
DestroyTree(pNode1);
}
// 树中只有1个结点
void Test4()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
Test("Test4", pNode1);
DestroyTree(pNode1);
}
// 树中没有结点
void Test5()
{
Test("Test5", nullptr);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
return 0;
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题32（二）：分行从上到下打印二叉树
// 题目：从上到下按层打印二叉树，同一层的结点按从左到右的顺序打印，每一层
// 打印到一行。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
#include <queue>
void Print(BinaryTreeNode* pRoot)
{
if(pRoot == nullptr)
return;
std::queue<BinaryTreeNode*> nodes;
nodes.push(pRoot);
int nextLevel = 0;
int toBePrinted = 1;
while(!nodes.empty())
{
BinaryTreeNode* pNode = nodes.front();
printf("%d ", pNode->m_nValue);
if(pNode->m_pLeft != nullptr)
{
nodes.push(pNode->m_pLeft);
++nextLevel;
}
if(pNode->m_pRight != nullptr)
{
nodes.push(pNode->m_pRight);
++nextLevel;
}
nodes.pop();
--toBePrinted;
if(toBePrinted == 0)
{
printf("\n");
toBePrinted = nextLevel;
nextLevel = 0;
}
}
}
// ====================测试代码====================
// 8
// 6 10
// 5 7 9 11
void Test1()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);
ConnectTreeNodes(pNode8, pNode6, pNode10);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);
printf("====Test1 Begins: ====\n");
printf("Expected Result is:\n");
printf("8 \n");
printf("6 10 \n");
printf("5 7 9 11 \n\n");
printf("Actual Result is: \n");
Print(pNode8);
printf("\n");
DestroyTree(pNode8);
}
// 5
// 4
// 3
// 2
void Test2()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
printf("====Test2 Begins: ====\n");
printf("Expected Result is:\n");
printf("5 \n");
printf("4 \n");
printf("3 \n");
printf("2 \n\n");
printf("Actual Result is: \n");
Print(pNode5);
printf("\n");
DestroyTree(pNode5);
}
// 5
// 4
// 3
// 2
void Test3()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNode5, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode2);
printf("====Test3 Begins: ====\n");
printf("Expected Result is:\n");
printf("5 \n");
printf("4 \n");
printf("3 \n");
printf("2 \n\n");
printf("Actual Result is: \n");
Print(pNode5);
printf("\n");
DestroyTree(pNode5);
}
void Test4()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
printf("====Test4 Begins: ====\n");
printf("Expected Result is:\n");
printf("5 \n\n");
printf("Actual Result is: \n");
Print(pNode5);
printf("\n");
DestroyTree(pNode5);
}
void Test5()
{
printf("====Test5 Begins: ====\n");
printf("Expected Result is:\n");
printf("Actual Result is: \n");
Print(nullptr);
printf("\n");
}
// 100
// /
// 50
// \
// 150
void Test6()
{
BinaryTreeNode* pNode100 = CreateBinaryTreeNode(100);
BinaryTreeNode* pNode50 = CreateBinaryTreeNode(50);
BinaryTreeNode* pNode150 = CreateBinaryTreeNode(150);
ConnectTreeNodes(pNode100, pNode50, nullptr);
ConnectTreeNodes(pNode50, nullptr, pNode150);
printf("====Test6 Begins: ====\n");
printf("Expected Result is:\n");
printf("100 \n");
printf("50 \n");
printf("150 \n\n");
printf("Actual Result is: \n");
Print(pNode100);
printf("\n");
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
return 0;
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题32（三）：之字形打印二叉树
// 题目：请实现一个函数按照之字形顺序打印二叉树，即第一行按照从左到右的顺
// 序打印，第二层按照从右到左的顺序打印，第三行再按照从左到右的顺序打印，
// 其他行以此类推。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
#include <stack>
void Print(BinaryTreeNode* pRoot)
{
if(pRoot == nullptr)
return;
std::stack<BinaryTreeNode*> levels[2];
int current = 0;
int next = 1;
levels[current].push(pRoot);
while(!levels[0].empty() || !levels[1].empty())
{
BinaryTreeNode* pNode = levels[current].top();
levels[current].pop();
printf("%d ", pNode->m_nValue);
if(current == 0)
{
if(pNode->m_pLeft != nullptr)
levels[next].push(pNode->m_pLeft);
if(pNode->m_pRight != nullptr)
levels[next].push(pNode->m_pRight);
}
else
{
if(pNode->m_pRight != nullptr)
levels[next].push(pNode->m_pRight);
if(pNode->m_pLeft != nullptr)
levels[next].push(pNode->m_pLeft);
}
if(levels[current].empty())
{
printf("\n");
current = 1 - current;
next = 1 - next;
}
}
}
// ====================测试代码====================
// 8
// 6 10
// 5 7 9 11
void Test1()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);
ConnectTreeNodes(pNode8, pNode6, pNode10);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);
printf("====Test1 Begins: ====\n");
printf("Expected Result is:\n");
printf("8 \n");
printf("10 6 \n");
printf("5 7 9 11 \n\n");
printf("Actual Result is: \n");
Print(pNode8);
printf("\n");
DestroyTree(pNode8);
}
// 5
// 4
// 3
// 2
void Test2()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
printf("====Test2 Begins: ====\n");
printf("Expected Result is:\n");
printf("5 \n");
printf("4 \n");
printf("3 \n");
printf("2 \n\n");
printf("Actual Result is: \n");
Print(pNode5);
printf("\n");
DestroyTree(pNode5);
}
// 5
// 4
// 3
// 2
void Test3()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNode5, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode2);
printf("====Test3 Begins: ====\n");
printf("Expected Result is:\n");
printf("5 \n");
printf("4 \n");
printf("3 \n");
printf("2 \n\n");
printf("Actual Result is: \n");
Print(pNode5);
printf("\n");
DestroyTree(pNode5);
}
void Test4()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
printf("====Test4 Begins: ====\n");
printf("Expected Result is:\n");
printf("5 \n\n");
printf("Actual Result is: \n");
Print(pNode5);
printf("\n");
DestroyTree(pNode5);
}
void Test5()
{
printf("====Test5 Begins: ====\n");
printf("Expected Result is:\n");
printf("Actual Result is: \n");
Print(nullptr);
printf("\n");
}
// 100
// /
// 50
// \
// 150
void Test6()
{
BinaryTreeNode* pNode100 = CreateBinaryTreeNode(100);
BinaryTreeNode* pNode50 = CreateBinaryTreeNode(50);
BinaryTreeNode* pNode150 = CreateBinaryTreeNode(150);
ConnectTreeNodes(pNode100, pNode50, nullptr);
ConnectTreeNodes(pNode50, nullptr, pNode150);
printf("====Test6 Begins: ====\n");
printf("Expected Result is:\n");
printf("100 \n");
printf("50 \n");
printf("150 \n\n");
printf("Actual Result is: \n");
Print(pNode100);
printf("\n");
}
// 8
// 4 12
// 2 6 10 14
// 1 3 5 7 9 11 13 15
void Test7()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode12 = CreateBinaryTreeNode(12);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode14 = CreateBinaryTreeNode(14);
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);
BinaryTreeNode* pNode13 = CreateBinaryTreeNode(13);
BinaryTreeNode* pNode15 = CreateBinaryTreeNode(15);
ConnectTreeNodes(pNode8, pNode4, pNode12);
ConnectTreeNodes(pNode4, pNode2, pNode6);
ConnectTreeNodes(pNode12, pNode10, pNode14);
ConnectTreeNodes(pNode2, pNode1, pNode3);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);
ConnectTreeNodes(pNode14, pNode13, pNode15);
printf("====Test7 Begins: ====\n");
printf("Expected Result is:\n");
printf("8 \n");
printf("12 4 \n");
printf("2 6 10 14 \n");
printf("15 13 11 9 7 5 3 1 \n\n");
printf("Actual Result is: \n");
Print(pNode8);
printf("\n");
DestroyTree(pNode8);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
return 0;
}
33. 面试题33：二叉搜索树的后序遍历序列
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题33：二叉搜索树的后序遍历序列
// 题目：输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历的结果。
// 如果是则返回true，否则返回false。假设输入的数组的任意两个数字都互不相同。
#include <cstdio>
// BST：Binary Search Tree，二叉搜索树
bool VerifySquenceOfBST(int sequence[], int length)
{
if(sequence == nullptr || length <= 0)
return false;
int root = sequence[length - 1];
// 在二叉搜索树中左子树的结点小于根结点
int i = 0;
for(; i < length - 1; ++ i)
{
if(sequence[i] > root)
break;
}
// 在二叉搜索树中右子树的结点大于根结点
int j = i;
for(; j < length - 1; ++ j)
{
if(sequence[j] < root)
return false;
}
// 判断左子树是不是二叉搜索树
bool left = true;
if(i > 0)
left = VerifySquenceOfBST(sequence, i);
// 判断右子树是不是二叉搜索树
bool right = true;
if(i < length - 1)
right = VerifySquenceOfBST(sequence + i, length - i - 1);
return (left && right);
}
// ====================测试代码====================
void Test(const char* testName, int sequence[], int length, bool expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
if(VerifySquenceOfBST(sequence, length) == expected)
printf("passed.\n");
else
printf("failed.\n");
}
// 10
// / \
// 6 14
// /\ /\
// 4 8 12 16
void Test1()
{
int data[] = {4, 8, 6, 12, 16, 14, 10};
Test("Test1", data, sizeof(data)/sizeof(int), true);
}
// 5
// / \
// 4 7
// /
// 6
void Test2()
{
int data[] = {4, 6, 7, 5};
Test("Test2", data, sizeof(data)/sizeof(int), true);
}
// 5
// /
// 4
// /
// 3
// /
// 2
// /
// 1
void Test3()
{
int data[] = {1, 2, 3, 4, 5};
Test("Test3", data, sizeof(data)/sizeof(int), true);
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void Test4()
{
int data[] = {5, 4, 3, 2, 1};
Test("Test4", data, sizeof(data)/sizeof(int), true);
}
// 树中只有1个结点
void Test5()
{
int data[] = {5};
Test("Test5", data, sizeof(data)/sizeof(int), true);
}
void Test6()
{
int data[] = {7, 4, 6, 5};
Test("Test6", data, sizeof(data)/sizeof(int), false);
}
void Test7()
{
int data[] = {4, 6, 12, 8, 16, 14, 10};
Test("Test7", data, sizeof(data)/sizeof(int), false);
}
void Test8()
{
Test("Test8", nullptr, 0, false);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
Test8();
return 0;
}
34. 面试题34：二叉树中和为某一值的路径
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题34：二叉树中和为某一值的路径
// 题目：输入一棵二叉树和一个整数，打印出二叉树中结点值的和为输入整数的所
// 有路径。从树的根结点开始往下一直到叶结点所经过的结点形成一条路径。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
#include <vector>
void FindPath(BinaryTreeNode* pRoot, int expectedSum, std::vector<int>& path, int& currentSum);
void FindPath(BinaryTreeNode* pRoot, int expectedSum)
{
if(pRoot == nullptr)
return;
std::vector<int> path;
int currentSum = 0;
FindPath(pRoot, expectedSum, path, currentSum);
}
void FindPath
(
BinaryTreeNode* pRoot,
int expectedSum,
std::vector<int>& path,
int& currentSum
)
{
currentSum += pRoot->m_nValue;
path.push_back(pRoot->m_nValue);
// 如果是叶结点，并且路径上结点的和等于输入的值
// 打印出这条路径
bool isLeaf = pRoot->m_pLeft == nullptr && pRoot->m_pRight == nullptr;
if(currentSum == expectedSum && isLeaf)
{
printf("A path is found: ");
std::vector<int>::iterator iter = path.begin();
for(; iter != path.end(); ++ iter)
printf("%d\t", *iter);
printf("\n");
}
// 如果不是叶结点，则遍历它的子结点
if(pRoot->m_pLeft != nullptr)
FindPath(pRoot->m_pLeft, expectedSum, path, currentSum);
if(pRoot->m_pRight != nullptr)
FindPath(pRoot->m_pRight, expectedSum, path, currentSum);
// 在返回到父结点之前，在路径上删除当前结点，
// 并在currentSum中减去当前结点的值
currentSum -= pRoot->m_nValue;
path.pop_back();
}
// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pRoot, int expectedSum)
{
if(testName != nullptr)
printf("%s begins:\n", testName);
FindPath(pRoot, expectedSum);
printf("\n");
}
// 10
// / \
// 5 12
// /\
// 4 7
// 有两条路径上的结点和为22
void Test1()
{
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode12 = CreateBinaryTreeNode(12);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNode10, pNode5, pNode12);
ConnectTreeNodes(pNode5, pNode4, pNode7);
printf("Two paths should be found in Test1.\n");
Test("Test1", pNode10, 22);
DestroyTree(pNode10);
}
// 10
// / \
// 5 12
// /\
// 4 7
// 没有路径上的结点和为15
void Test2()
{
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode12 = CreateBinaryTreeNode(12);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNode10, pNode5, pNode12);
ConnectTreeNodes(pNode5, pNode4, pNode7);
printf("No paths should be found in Test2.\n");
Test("Test2", pNode10, 15);
DestroyTree(pNode10);
}
// 5
// /
// 4
// /
// 3
// /
// 2
// /
// 1
// 有一条路径上面的结点和为15
void Test3()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
ConnectTreeNodes(pNode2, pNode1, nullptr);
printf("One path should be found in Test3.\n");
Test("Test3", pNode5, 15);
DestroyTree(pNode5);
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
// 没有路径上面的结点和为16
void Test4()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, nullptr, pNode2);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
printf("No paths should be found in Test4.\n");
Test("Test4", pNode1, 16);
DestroyTree(pNode1);
}
// 树中只有1个结点
void Test5()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
printf("One path should be found in Test5.\n");
Test("Test5", pNode1, 1);
DestroyTree(pNode1);
}
// 树中没有结点
void Test6()
{
printf("No paths should be found in Test6.\n");
Test("Test6", nullptr, 0);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
return 0;
}
36. 面试题36：二叉搜索树与双向链表
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题36：二叉搜索树与双向链表
// 题目：输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求
// 不能创建任何新的结点，只能调整树中结点指针的指向。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
void ConvertNode(BinaryTreeNode* pNode, BinaryTreeNode** pLastNodeInList);
BinaryTreeNode* Convert(BinaryTreeNode* pRootOfTree)
{
BinaryTreeNode *pLastNodeInList = nullptr;
ConvertNode(pRootOfTree, &pLastNodeInList);
// pLastNodeInList指向双向链表的尾结点，
// 我们需要返回头结点
BinaryTreeNode *pHeadOfList = pLastNodeInList;
while(pHeadOfList != nullptr && pHeadOfList->m_pLeft != nullptr)
pHeadOfList = pHeadOfList->m_pLeft;
return pHeadOfList;
}
void ConvertNode(BinaryTreeNode* pNode, BinaryTreeNode** pLastNodeInList)
{
if(pNode == nullptr)
return;
BinaryTreeNode *pCurrent = pNode;
if (pCurrent->m_pLeft != nullptr)
ConvertNode(pCurrent->m_pLeft, pLastNodeInList);
pCurrent->m_pLeft = *pLastNodeInList;
if(*pLastNodeInList != nullptr)
(*pLastNodeInList)->m_pRight = pCurrent;
*pLastNodeInList = pCurrent;
if (pCurrent->m_pRight != nullptr)
ConvertNode(pCurrent->m_pRight, pLastNodeInList);
}
// ====================测试代码====================
void PrintDoubleLinkedList(BinaryTreeNode* pHeadOfList)
{
BinaryTreeNode* pNode = pHeadOfList;
printf("The nodes from left to right are:\n");
while(pNode != nullptr)
{
printf("%d\t", pNode->m_nValue);
if(pNode->m_pRight == nullptr)
break;
pNode = pNode->m_pRight;
}
printf("\nThe nodes from right to left are:\n");
while(pNode != nullptr)
{
printf("%d\t", pNode->m_nValue);
if(pNode->m_pLeft == nullptr)
break;
pNode = pNode->m_pLeft;
}
printf("\n");
}
void DestroyList(BinaryTreeNode* pHeadOfList)
{
BinaryTreeNode* pNode = pHeadOfList;
while(pNode != nullptr)
{
BinaryTreeNode* pNext = pNode->m_pRight;
delete pNode;
pNode = pNext;
}
}
void Test(char* testName, BinaryTreeNode* pRootOfTree)
{
if(testName != nullptr)
printf("%s begins:\n", testName);
PrintTree(pRootOfTree);
BinaryTreeNode* pHeadOfList = Convert(pRootOfTree);
PrintDoubleLinkedList(pHeadOfList);
}
// 10
// / \
// 6 14
// /\ /\
// 4 8 12 16
void Test1()
{
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode14 = CreateBinaryTreeNode(14);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode12 = CreateBinaryTreeNode(12);
BinaryTreeNode* pNode16 = CreateBinaryTreeNode(16);
ConnectTreeNodes(pNode10, pNode6, pNode14);
ConnectTreeNodes(pNode6, pNode4, pNode8);
ConnectTreeNodes(pNode14, pNode12, pNode16);
Test("Test1", pNode10);
DestroyList(pNode4);
}
// 5
// /
// 4
// /
// 3
// /
// 2
// /
// 1
void Test2()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
ConnectTreeNodes(pNode2, pNode1, nullptr);
Test("Test2", pNode5);
DestroyList(pNode1);
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void Test3()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, nullptr, pNode2);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
Test("Test3", pNode1);
DestroyList(pNode1);
}
// 树中只有1个结点
void Test4()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
Test("Test4", pNode1);
DestroyList(pNode1);
}
// 树中没有结点
void Test5()
{
Test("Test5", nullptr);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
return 0;
}
37. 面试题37：序列化二叉树
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题37：序列化二叉树
// 题目：请实现两个函数，分别用来序列化和反序列化二叉树。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
#include <iostream>
#include <fstream>
using namespace std;
void Serialize(const BinaryTreeNode* pRoot, ostream& stream)
{
if(pRoot == nullptr)
{
stream << "$,";
return;
}
stream << pRoot->m_nValue << ',';
Serialize(pRoot->m_pLeft, stream);
Serialize(pRoot->m_pRight, stream);
}
bool ReadStream(istream& stream, int* number)
{
if(stream.eof())
return false;
char buffer[32];
buffer[0] = '\0';
char ch;
stream >> ch;
int i = 0;
while(!stream.eof() && ch != ',')
{
buffer[i++] = ch;
stream >> ch;
}
bool isNumeric = false;
if(i > 0 && buffer[0] != '$')
{
*number = atoi(buffer);
isNumeric = true;
}
return isNumeric;
}
void Deserialize(BinaryTreeNode** pRoot, istream& stream)
{
int number;
if(ReadStream(stream, &number))
{
*pRoot = new BinaryTreeNode();
(*pRoot)->m_nValue = number;
(*pRoot)->m_pLeft = nullptr;
(*pRoot)->m_pRight = nullptr;
Deserialize(&((*pRoot)->m_pLeft), stream);
Deserialize(&((*pRoot)->m_pRight), stream);
}
}
// ==================== Test Code ====================
bool isSameTree(const BinaryTreeNode* pRoot1, const BinaryTreeNode* pRoot2)
{
if(pRoot1 == nullptr && pRoot2 == nullptr)
return true;
if(pRoot1 == nullptr || pRoot2 == nullptr)
return false;
if(pRoot1->m_nValue != pRoot2->m_nValue)
return false;
return isSameTree(pRoot1->m_pLeft, pRoot2->m_pLeft) &&
isSameTree(pRoot1->m_pRight, pRoot2->m_pRight);
}
void Test(const char* testName, const BinaryTreeNode* pRoot)
{
if(testName != nullptr)
printf("%s begins: \n", testName);
PrintTree(pRoot);
char* fileName = "test.txt";
ofstream fileOut;
fileOut.open(fileName);
Serialize(pRoot, fileOut);
fileOut.close();
// print the serialized file
ifstream fileIn1;
char ch;
fileIn1.open(fileName);
while(!fileIn1.eof())
{
fileIn1 >> ch;
cout << ch;
}
fileIn1.close();
cout << endl;
ifstream fileIn2;
fileIn2.open(fileName);
BinaryTreeNode* pNewRoot = nullptr;
Deserialize(&pNewRoot, fileIn2);
fileIn2.close();
PrintTree(pNewRoot);
if(isSameTree(pRoot, pNewRoot))
printf("The deserialized tree is same as the oritinal tree.\n\n");
else
printf("The deserialized tree is NOT same as the oritinal tree.\n\n");
DestroyTree(pNewRoot);
}
// 8
// 6 10
// 5 7 9 11
void Test1()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);
ConnectTreeNodes(pNode8, pNode6, pNode10);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);
Test("Test1", pNode8);
DestroyTree(pNode8);
}
// 5
// 4
// 3
// 2
void Test2()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
Test("Test2", pNode5);
DestroyTree(pNode5);
}
// 5
// 4
// 3
// 2
void Test3()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
ConnectTreeNodes(pNode5, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode2);
Test("Test3", pNode5);
DestroyTree(pNode5);
}
void Test4()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
Test("Test4", pNode5);
DestroyTree(pNode5);
}
void Test5()
{
Test("Test5", nullptr);
}
// 5
// 5
// 5
// 5
// 5
// 5 5
// 5 5
void Test6()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode61 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode62 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode71 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode72 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, nullptr, pNode2);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode5, nullptr);
ConnectTreeNodes(pNode5, pNode61, pNode62);
ConnectTreeNodes(pNode61, pNode71, nullptr);
ConnectTreeNodes(pNode62, nullptr, pNode72);
Test("Test6", pNode1);
DestroyTree(pNode1);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
return 0;
}
54. 面试题54：二叉搜索树的第k个结点
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题54：二叉搜索树的第k个结点
// 题目：给定一棵二叉搜索树，请找出其中的第k大的结点。
#include <cstdio>
#include "../Utilities/BinaryTree.h"
const BinaryTreeNode* KthNodeCore(const BinaryTreeNode* pRoot, unsigned int& k);
const BinaryTreeNode* KthNode(const BinaryTreeNode* pRoot, unsigned int k)
{
if(pRoot == nullptr || k == 0)
return nullptr;
return KthNodeCore(pRoot, k);
}
const BinaryTreeNode* KthNodeCore(const BinaryTreeNode* pRoot, unsigned int& k)
{
const BinaryTreeNode* target = nullptr;
if(pRoot->m_pLeft != nullptr)
target = KthNodeCore(pRoot->m_pLeft, k);
if(target == nullptr)
{
if(k == 1)
target = pRoot;
k--;
}
if(target == nullptr && pRoot->m_pRight != nullptr)
target = KthNodeCore(pRoot->m_pRight, k);
return target;
}
// ====================测试代码====================
void Test(const char* testName, const BinaryTreeNode* pRoot, unsigned int k, bool isNull, int expected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
const BinaryTreeNode* pTarget = KthNode(pRoot, k);
if((isNull && pTarget == nullptr) || (!isNull && pTarget->m_nValue == expected))
printf("Passed.\n");
else
printf("FAILED.\n");
}
// 8
// 6 10
// 5 7 9 11
void TestA()
{
BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
BinaryTreeNode* pNode9 = CreateBinaryTreeNode(9);
BinaryTreeNode* pNode11 = CreateBinaryTreeNode(11);
ConnectTreeNodes(pNode8, pNode6, pNode10);
ConnectTreeNodes(pNode6, pNode5, pNode7);
ConnectTreeNodes(pNode10, pNode9, pNode11);
Test("TestA0", pNode8, 0, true, -1);
Test("TestA1", pNode8, 1, false, 5);
Test("TestA2", pNode8, 2, false, 6);
Test("TestA3", pNode8, 3, false, 7);
Test("TestA4", pNode8, 4, false, 8);
Test("TestA5", pNode8, 5, false, 9);
Test("TestA6", pNode8, 6, false, 10);
Test("TestA7", pNode8, 7, false, 11);
Test("TestA8", pNode8, 8, true, -1);
DestroyTree(pNode8);
printf("\n\n");
}
// 5
// /
// 4
// /
// 3
// /
// 2
// /
// 1
void TestB()
{
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
ConnectTreeNodes(pNode5, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode2, nullptr);
ConnectTreeNodes(pNode2, pNode1, nullptr);
Test("TestB0", pNode5, 0, true, -1);
Test("TestB1", pNode5, 1, false, 1);
Test("TestB2", pNode5, 2, false, 2);
Test("TestB3", pNode5, 3, false, 3);
Test("TestB4", pNode5, 4, false, 4);
Test("TestB5", pNode5, 5, false, 5);
Test("TestB6", pNode5, 6, true, -1);
DestroyTree(pNode5);
printf("\n\n");
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void TestC()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, nullptr, pNode2);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
Test("TestC0", pNode1, 0, true, -1);
Test("TestC1", pNode1, 1, false, 1);
Test("TestC2", pNode1, 2, false, 2);
Test("TestC3", pNode1, 3, false, 3);
Test("TestC4", pNode1, 4, false, 4);
Test("TestC5", pNode1, 5, false, 5);
Test("TestC6", pNode1, 6, true, -1);
DestroyTree(pNode1);
printf("\n\n");
}
// There is only one node in a tree
void TestD()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
Test("TestD0", pNode1, 0, true, -1);
Test("TestD1", pNode1, 1, false, 1);
Test("TestD2", pNode1, 2, true, -1);
DestroyTree(pNode1);
printf("\n\n");
}
// empty tree
void TestE()
{
Test("TestE0", nullptr, 0, true, -1);
Test("TestE1", nullptr, 1, true, -1);
printf("\n\n");
}
int main(int argc, char* argv[])
{
TestA();
TestB();
TestC();
TestD();
TestE();
}
55. 面试题55：二叉树的深度
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题55（一）：二叉树的深度
// 题目：输入一棵二叉树的根结点，求该树的深度。从根结点到叶结点依次经过的
// 结点（含根、叶结点）形成树的一条路径，最长路径的长度为树的深度。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
int TreeDepth(const BinaryTreeNode* pRoot)
{
if(pRoot == nullptr)
return 0;
int nLeft = TreeDepth(pRoot->m_pLeft);
int nRight = TreeDepth(pRoot->m_pRight);
return (nLeft > nRight) ? (nLeft + 1) : (nRight + 1);
}
// ====================测试代码====================
void Test(const char* testName, const BinaryTreeNode* pRoot, int expected)
{
int result = TreeDepth(pRoot);
if(expected == result)
printf("%s passed.\n", testName);
else
printf("%s FAILED.\n", testName);
}
// 1
// / \
// 2 3
// /\ \
// 4 5 6
// /
// 7
void Test1()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNode1, pNode2, pNode3);
ConnectTreeNodes(pNode2, pNode4, pNode5);
ConnectTreeNodes(pNode3, nullptr, pNode6);
ConnectTreeNodes(pNode5, pNode7, nullptr);
Test("Test1", pNode1, 4);
DestroyTree(pNode1);
}
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
void Test2()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, pNode2, nullptr);
ConnectTreeNodes(pNode2, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode5, nullptr);
Test("Test2", pNode1, 5);
DestroyTree(pNode1);
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void Test3()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, nullptr, pNode2);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
Test("Test3", pNode1, 5);
DestroyTree(pNode1);
}
// 树中只有1个结点
void Test4()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
Test("Test4", pNode1, 1);
DestroyTree(pNode1);
}
// 树中没有结点
void Test5()
{
Test("Test5", nullptr, 0);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
return 0;
}
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题55（二）：平衡二叉树
// 题目：输入一棵二叉树的根结点，判断该树是不是平衡二叉树。如果某二叉树中
// 任意结点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。
#include <cstdio>
#include "..\Utilities\BinaryTree.h"
// ====================方法1====================
int TreeDepth(const BinaryTreeNode* pRoot)
{
if(pRoot == nullptr)
return 0;
int nLeft = TreeDepth(pRoot->m_pLeft);
int nRight = TreeDepth(pRoot->m_pRight);
return (nLeft > nRight) ? (nLeft + 1) : (nRight + 1);
}
bool IsBalanced_Solution1(const BinaryTreeNode* pRoot)
{
if(pRoot == nullptr)
return true;
int left = TreeDepth(pRoot->m_pLeft);
int right = TreeDepth(pRoot->m_pRight);
int diff = left - right;
if(diff > 1 || diff < -1)
return false;
return IsBalanced_Solution1(pRoot->m_pLeft)
&& IsBalanced_Solution1(pRoot->m_pRight);
}
// ====================方法2====================
bool IsBalanced(const BinaryTreeNode* pRoot, int* pDepth);
bool IsBalanced_Solution2(const BinaryTreeNode* pRoot)
{
int depth = 0;
return IsBalanced(pRoot, &depth);
}
bool IsBalanced(const BinaryTreeNode* pRoot, int* pDepth)
{
if(pRoot == nullptr)
{
*pDepth = 0;
return true;
}
int left, right;
if(IsBalanced(pRoot->m_pLeft, &left)
&& IsBalanced(pRoot->m_pRight, &right))
{
int diff = left - right;
if(diff <= 1 && diff >= -1)
{
*pDepth = 1 + (left > right ? left : right);
return true;
}
}
return false;
}
// ====================测试代码====================
void Test(const char* testName, const BinaryTreeNode* pRoot, bool expected)
{
if(testName != nullptr)
printf("%s begins:\n", testName);
printf("Solution1 begins: ");
if(IsBalanced_Solution1(pRoot) == expected)
printf("Passed.\n");
else
printf("Failed.\n");
printf("Solution2 begins: ");
if(IsBalanced_Solution2(pRoot) == expected)
printf("Passed.\n");
else
printf("Failed.\n");
}
// 完全二叉树
// 1
// / \
// 2 3
// /\ / \
// 4 5 6 7
void Test1()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNode1, pNode2, pNode3);
ConnectTreeNodes(pNode2, pNode4, pNode5);
ConnectTreeNodes(pNode3, pNode6, pNode7);
Test("Test1", pNode1, true);
DestroyTree(pNode1);
}
// 不是完全二叉树，但是平衡二叉树
// 1
// / \
// 2 3
// /\ \
// 4 5 6
// /
// 7
void Test2()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
BinaryTreeNode* pNode7 = CreateBinaryTreeNode(7);
ConnectTreeNodes(pNode1, pNode2, pNode3);
ConnectTreeNodes(pNode2, pNode4, pNode5);
ConnectTreeNodes(pNode3, nullptr, pNode6);
ConnectTreeNodes(pNode5, pNode7, nullptr);
Test("Test2", pNode1, true);
DestroyTree(pNode1);
}
// 不是平衡二叉树
// 1
// / \
// 2 3
// /\
// 4 5
// /
// 6
void Test3()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
ConnectTreeNodes(pNode1, pNode2, pNode3);
ConnectTreeNodes(pNode2, pNode4, pNode5);
ConnectTreeNodes(pNode5, pNode6, nullptr);
Test("Test3", pNode1, false);
DestroyTree(pNode1);
}
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
void Test4()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, pNode2, nullptr);
ConnectTreeNodes(pNode2, pNode3, nullptr);
ConnectTreeNodes(pNode3, pNode4, nullptr);
ConnectTreeNodes(pNode4, pNode5, nullptr);
Test("Test4", pNode1, false);
DestroyTree(pNode1);
}
// 1
// \
// 2
// \
// 3
// \
// 4
// \
// 5
void Test5()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
ConnectTreeNodes(pNode1, nullptr, pNode2);
ConnectTreeNodes(pNode2, nullptr, pNode3);
ConnectTreeNodes(pNode3, nullptr, pNode4);
ConnectTreeNodes(pNode4, nullptr, pNode5);
Test("Test5", pNode1, false);
DestroyTree(pNode1);
}
// 树中只有1个结点
void Test6()
{
BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
Test("Test6", pNode1, true);
DestroyTree(pNode1);
}
// 树中没有结点
void Test7()
{
Test("Test7", nullptr, true);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
Test5();
Test6();
Test7();
return 0;
}
68. 面试题68：树中两个结点的最低公共祖先
/*******************************************************************
Copyright(c) 2016, Harry He
All rights reserved.
Distributed under the BSD license.
(See accompanying file LICENSE.txt at
https://github.com/zhedahht/CodingInterviewChinese2/blob/master/LICENSE.txt)
*******************************************************************/
//==================================================================
// 《剑指Offer——名企面试官精讲典型编程题》代码
// 作者：何海涛
//==================================================================
// 面试题68：树中两个结点的最低公共祖先
// 题目：输入两个树结点，求它们的最低公共祖先。
#include <cstdio>
#include "..\Utilities\Tree.h"
#include <list>
using namespace std;
bool GetNodePath(const TreeNode* pRoot, const TreeNode* pNode, list<const TreeNode*>& path)
{
if(pRoot == pNode)
return true;
path.push_back(pRoot);
bool found = false;
vector<TreeNode*>::const_iterator i = pRoot->m_vChildren.begin();
while(!found && i < pRoot->m_vChildren.end())
{
found = GetNodePath(*i, pNode, path);
++i;
}
if(!found)
path.pop_back();
return found;
}
const TreeNode* GetLastCommonNode
(
const list<const TreeNode*>& path1,
const list<const TreeNode*>& path2
)
{
list<const TreeNode*>::const_iterator iterator1 = path1.begin();
list<const TreeNode*>::const_iterator iterator2 = path2.begin();
const TreeNode* pLast = nullptr;
while(iterator1 != path1.end() && iterator2 != path2.end())
{
if(*iterator1 == *iterator2)
pLast = *iterator1;
iterator1++;
iterator2++;
}
return pLast;
}
const TreeNode* GetLastCommonParent(const TreeNode* pRoot, const TreeNode* pNode1, const TreeNode* pNode2)
{
if(pRoot == nullptr || pNode1 == nullptr || pNode2 == nullptr)
return nullptr;
list<const TreeNode*> path1;
GetNodePath(pRoot, pNode1, path1);
list<const TreeNode*> path2;
GetNodePath(pRoot, pNode2, path2);
return GetLastCommonNode(path1, path2);
}
// ====================测试代码====================
void Test(const char* testName, const TreeNode* pRoot, const TreeNode* pNode1, const TreeNode* pNode2, TreeNode* pExpected)
{
if(testName != nullptr)
printf("%s begins: ", testName);
const TreeNode* pResult = GetLastCommonParent(pRoot, pNode1, pNode2);
if((pExpected == nullptr && pResult == nullptr) ||
(pExpected != nullptr && pResult != nullptr && pResult->m_nValue == pExpected->m_nValue))
printf("Passed.\n");
else
printf("Failed.\n");
}
// 形状普通的树
// 1
// / \
// 2 3
// / \
// 4 5
// / \ / | \
// 6 7 8 9 10
void Test1()
{
TreeNode* pNode1 = CreateTreeNode(1);
TreeNode* pNode2 = CreateTreeNode(2);
TreeNode* pNode3 = CreateTreeNode(3);
TreeNode* pNode4 = CreateTreeNode(4);
TreeNode* pNode5 = CreateTreeNode(5);
TreeNode* pNode6 = CreateTreeNode(6);
TreeNode* pNode7 = CreateTreeNode(7);
TreeNode* pNode8 = CreateTreeNode(8);
TreeNode* pNode9 = CreateTreeNode(9);
TreeNode* pNode10 = CreateTreeNode(10);
ConnectTreeNodes(pNode1, pNode2);
ConnectTreeNodes(pNode1, pNode3);
ConnectTreeNodes(pNode2, pNode4);
ConnectTreeNodes(pNode2, pNode5);
ConnectTreeNodes(pNode4, pNode6);
ConnectTreeNodes(pNode4, pNode7);
ConnectTreeNodes(pNode5, pNode8);
ConnectTreeNodes(pNode5, pNode9);
ConnectTreeNodes(pNode5, pNode10);
Test("Test1", pNode1, pNode6, pNode8, pNode2);
}
// 树退化成一个链表
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
void Test2()
{
TreeNode* pNode1 = CreateTreeNode(1);
TreeNode* pNode2 = CreateTreeNode(2);
TreeNode* pNode3 = CreateTreeNode(3);
TreeNode* pNode4 = CreateTreeNode(4);
TreeNode* pNode5 = CreateTreeNode(5);
ConnectTreeNodes(pNode1, pNode2);
ConnectTreeNodes(pNode2, pNode3);
ConnectTreeNodes(pNode3, pNode4);
ConnectTreeNodes(pNode4, pNode5);
Test("Test2", pNode1, pNode5, pNode4, pNode3);
}
// 树退化成一个链表，一个结点不在树中
// 1
// /
// 2
// /
// 3
// /
// 4
// /
// 5
void Test3()
{
TreeNode* pNode1 = CreateTreeNode(1);
TreeNode* pNode2 = CreateTreeNode(2);
TreeNode* pNode3 = CreateTreeNode(3);
TreeNode* pNode4 = CreateTreeNode(4);
TreeNode* pNode5 = CreateTreeNode(5);
ConnectTreeNodes(pNode1, pNode2);
ConnectTreeNodes(pNode2, pNode3);
ConnectTreeNodes(pNode3, pNode4);
ConnectTreeNodes(pNode4, pNode5);
TreeNode* pNode6 = CreateTreeNode(6);
Test("Test3", pNode1, pNode5, pNode6, nullptr);
}
// 输入nullptr
void Test4()
{
Test("Test4", nullptr, nullptr, nullptr, nullptr);
}
int main(int argc, char* argv[])
{
Test1();
Test2();
Test3();
Test4();
return 0;
}
