112. 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.

判断是否存在一条从根节点到叶子节点的路径，路径上节点之和为sum

C++（6ms）：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool hasPathSum(TreeNode* root, int sum) {
13         if (root == NULL) return false ;
14         if (root->val == sum && root->left == NULL && root->right == NULL) return true ;
15         return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right,sum-root->val) ;
16     }
17 };