JZ-C-23

剑指offer第二十三题：从上往下打印二叉树

  1 //============================================================================
  2 // Name        : JZ-C-23.cpp
  3 // Author      : Laughing_Lz
  4 // Version     :
  5 // Copyright   : All Right Reserved
  6 // Description : 从上往下打印二叉树
  7 //============================================================================
  8 
  9 #include <iostream>
 10 #include <stdio.h>
 11 #include <deque>
 12 #include "BinaryTree.h"
 13 using namespace std;
 14 /**
 15  *层次遍历二叉树，相当于广度优先遍历,都需要利用队列
 16  */
 17 void PrintFromTopToBottom(BinaryTreeNode* pTreeRoot){
 18     if(pTreeRoot == NULL){
 19         return;
 20     }
 21     std::deque<BinaryTreeNode*> dequeTreeNode;//这是一个两端都可以进出的队列★
 22     dequeTreeNode.push_back(pTreeRoot);//从后进队列
 23     while(!dequeTreeNode.empty()){
 24         BinaryTreeNode* Node = dequeTreeNode.front();//从前出队列
 25         dequeTreeNode.pop_front();
 26         cout<<Node->m_nValue<<endl;//打印
 27         if(Node->m_pLeft){
 28             dequeTreeNode.push_back(Node->m_pLeft);
 29         }
 30         if(Node->m_pRight){
 31             dequeTreeNode.push_back(Node->m_pRight);
 32         }
 33     }
 34 }
 35 
 36 // ====================测试代码====================
 37 void Test(char* testName, BinaryTreeNode* pRoot)
 38 {
 39     if(testName != NULL)
 40         printf("%s begins: \n", testName);
 41 
 42     PrintTree(pRoot);
 43 
 44     printf("The nodes from top to bottom, from left to right are: \n");
 45     PrintFromTopToBottom(pRoot);
 46 
 47     printf("\n\n");
 48 }
 49 
 50 //            10
 51 //         /      \
 52 //        6        14
 53 //       /\        /\
 54 //      4  8     12  16
 55 void Test1()
 56 {
 57     BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
 58     BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
 59     BinaryTreeNode* pNode14 = CreateBinaryTreeNode(14);
 60     BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
 61     BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
 62     BinaryTreeNode* pNode12 = CreateBinaryTreeNode(12);
 63     BinaryTreeNode* pNode16 = CreateBinaryTreeNode(16);
 64 
 65     ConnectTreeNodes(pNode10, pNode6, pNode14);
 66     ConnectTreeNodes(pNode6, pNode4, pNode8);
 67     ConnectTreeNodes(pNode14, pNode12, pNode16);
 68 
 69     Test("Test1", pNode10);
 70 
 71     DestroyTree(pNode10);
 72 }
 73 
 74 //               5
 75 //              /
 76 //             4
 77 //            /
 78 //           3
 79 //          /
 80 //         2
 81 //        /
 82 //       1
 83 void Test2()
 84 {
 85     BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
 86     BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
 87     BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
 88     BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
 89     BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
 90 
 91     ConnectTreeNodes(pNode5, pNode4, NULL);
 92     ConnectTreeNodes(pNode4, pNode3, NULL);
 93     ConnectTreeNodes(pNode3, pNode2, NULL);
 94     ConnectTreeNodes(pNode2, pNode1, NULL);
 95 
 96     Test("Test2", pNode5);
 97 
 98     DestroyTree(pNode5);
 99 }
100 
101 // 1
102 //  \
103 //   2
104 //    \
105 //     3
106 //      \
107 //       4
108 //        \
109 //         5
110 void Test3()
111 {
112     BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
113     BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
114     BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
115     BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
116     BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
117 
118     ConnectTreeNodes(pNode1, NULL, pNode2);
119     ConnectTreeNodes(pNode2, NULL, pNode3);
120     ConnectTreeNodes(pNode3, NULL, pNode4);
121     ConnectTreeNodes(pNode4, NULL, pNode5);
122 
123     Test("Test3", pNode1);
124 
125     DestroyTree(pNode1);
126 }
127 
128 // 树中只有1个结点
129 void Test4()
130 {
131     BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
132     Test("Test4", pNode1);
133 
134     DestroyTree(pNode1);
135 }
136 
137 // 树中没有结点
138 void Test5()
139 {
140     Test("Test5", NULL);
141 }
142 
143 int main(int argc, char** argv)
144 {
145     Test1();
146     Test2();
147     Test3();
148     Test4();
149     Test5();
150 
151    return 0;
152 }

 

　　

posted @ 2016-06-13 15:00  回看欧洲  阅读(159)  评论(0)  编辑  收藏  举报