112. Path Sum

题目：

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.

思路：

本题依旧用递归实现，采用dfs，判断是否相等的临界条件：当前点为叶子结点。

代码：

/** * Definition for a binary tree node. * 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) {        return sumRes(root,sum,0);    }private:    bool sumRes(TreeNode* root,int sum,int cur){        if(root==NULL)            return false;        if(root->left==NULL&&root->right==NULL)            return (cur+root->val)==sum;        return sumRes(root->left,sum,cur+root->val)||sumRes(root->right,sum,cur+root->val);            }};