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.
解题思路：
1，若树为空返回false
2，若树只有一个根节点，返回 root->val == sum ?
3，若树没有左子树，在右子树中搜索
4，若树没有右子树，在左子树中搜索
5，否则，搜索左子树和右子树
代码：
1 /** 2 * Definition for binary tree 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 15 if (root->left == NULL && root->right == NULL && root->val == sum) return true; 16 17 if (root->left == NULL) return hasPathSum(root->right, sum - root->val); 18 else if (root->right == NULL) return hasPathSum(root->left, sum - root->val); 19 else return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); 20 } 21 };