Path Sum

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.

和前面的最小深度有点儿类似，只是要求对节点上的值做一些处理而已。每次递归都把sum减去当前节点的value就好了，如果递归到叶节点值刚好相等就return true,否则false。

/** * 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)return false;if (sum == root->val && !root->left && !root->right)return true;bool left = hasPathSum(root->left,sum-root->val);bool right = hasPathSum(root->right,sum-root->val);if (left || right)return true;return false;    }};