Path Sum
来源:互联网 发布:淘宝多少个差评封店 编辑:程序博客网 时间:2024/05/16 16:20
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:Given the below binary tree and
sum = 22
,5 / \ 4 8 / / \ 11 13 4 / \ \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: bool hasPathSum(TreeNode *root, int sum) { if(root){ sum-=root->val; }else{ return false; } if(NULL==root->left&&NULL==root->right){ return sum==0; } return hasPathSum(root->left, sum)||hasPathSum(root->right, sum); }};
0 0
- Path Sum && Path Sum ||
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- Path Sum
- 网络小工具积累
- 亲和数问题
- 7、XML处理
- 动态规划(DP)算法
- DAPM之一:概述
- Path Sum
- 技术博客
- 苹果的新编程语言 Swift 简介
- c++ 字符类型总结区别wchar_t,char,WCHAR(转)
- 完美洗牌算法
- 数据库SQL 约束
- Linux下的压缩解压缩命令详解
- HDOJ Train Problem II(Java)
- 修改在客户端显示“hello world”在server中是把几个块先合并成一个完整的数据再修