【二叉树的递归】07路径组成数字的和【Sum Root to Leaf Numbers】
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给定一个二叉树,节点的值仅限于从0-9,每一个从根节点达到叶子节点的路径代表一个数字。
一个例子,如果根节点到叶子节点的路径是 1->2->3,那么代表这个数字是123。
寻找所有路径代表的数字的和。
例如:
1 / \ 2 3
从根节点到叶子节点的路径是 1->2
代表的数字是 12
.从根节点到叶子节点的路径是 1->3
代表的数字是 13
.
返回和为 sum = 12 + 13 = 25
.
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Given a binary tree containing digits from 0-9
only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path 1->2->3
which represents the number 123
.
Find the total sum of all root-to-leaf numbers.
For example,
1 / \ 2 3
The root-to-leaf path 1->2
represents the number 12
.
The root-to-leaf path 1->3
represents the number 13
.
Return the sum = 12 + 13 = 25
.
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test.cpp:
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#include <iostream>
#include <cstdio> #include <stack> #include <vector> #include "BinaryTree.h" using namespace std; /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ int sumNumbers(TreeNode *root, int last) { /*没进入一层都要把last乘以10*/ if(root->left == NULL && root->right == NULL) { return last * 10 + root->val; } else if(root->left == NULL) { return sumNumbers(root->right, last * 10 + root->val); } else if(root->right == NULL) { return sumNumbers(root->left, last * 10 + root->val); } else { return sumNumbers(root->left, last * 10 + root->val) + sumNumbers(root->right, last * 10 + root->val); } } int sumNumbers(TreeNode *root) { /*空树*/ if(root == NULL) { return 0; } else { return sumNumbers(root, 0); } } // 树中结点含有分叉, // 8 // / \ // 6 1 // / \ // 9 2 // / \ // 4 7 int main() { TreeNode *pNodeA1 = CreateBinaryTreeNode(8); TreeNode *pNodeA2 = CreateBinaryTreeNode(6); TreeNode *pNodeA3 = CreateBinaryTreeNode(1); TreeNode *pNodeA4 = CreateBinaryTreeNode(9); TreeNode *pNodeA5 = CreateBinaryTreeNode(2); TreeNode *pNodeA6 = CreateBinaryTreeNode(4); TreeNode *pNodeA7 = CreateBinaryTreeNode(7); ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5); ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7); PrintTree(pNodeA1); //81 + 8627 + 8624 + 869 = 18201 cout << sumNumbers(pNodeA1) << endl; DestroyTree(pNodeA1); return 0; } |
结果输出:
18201
BinaryTree.h:
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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_ struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; TreeNode *CreateBinaryTreeNode(int value); void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight); void PrintTreeNode(TreeNode *pNode); void PrintTree(TreeNode *pRoot); void DestroyTree(TreeNode *pRoot); #endif /*_BINARY_TREE_H_*/ |
BinaryTree.cpp:
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#include <iostream>
#include <cstdio> #include "BinaryTree.h" using namespace std; /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ //创建结点 TreeNode *CreateBinaryTreeNode(int value) { TreeNode *pNode = new TreeNode(value); return pNode; } //连接结点 void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight) { if(pParent != NULL) { pParent->left = pLeft; pParent->right = pRight; } } //打印节点内容以及左右子结点内容 void PrintTreeNode(TreeNode *pNode) { if(pNode != NULL) { printf("value of this node is: %d\n", pNode->val); if(pNode->left != NULL) printf("value of its left child is: %d.\n", pNode->left->val); else printf("left child is null.\n"); if(pNode->right != NULL) printf("value of its right child is: %d.\n", pNode->right->val); else printf("right child is null.\n"); } else { printf("this node is null.\n"); } printf("\n"); } //前序遍历递归方法打印结点内容 void PrintTree(TreeNode *pRoot) { PrintTreeNode(pRoot); if(pRoot != NULL) { if(pRoot->left != NULL) PrintTree(pRoot->left); if(pRoot->right != NULL) PrintTree(pRoot->right); } } void DestroyTree(TreeNode *pRoot) { if(pRoot != NULL) { TreeNode *pLeft = pRoot->left; TreeNode *pRight = pRoot->right; delete pRoot; pRoot = NULL; DestroyTree(pLeft); DestroyTree(pRight); } } |