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.
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) { if (!root) return false; int next = sum - root->val; if ( !root->left && !root->right ) return sum==root->val; if ( root->left && !root->right ) return Solution::hasPathSum(root->left, next); if ( !root->left && root->right ) return Solution::hasPathSum(root->right, next); return Solution::hasPathSum(root->left, next) || Solution::hasPathSum(root->right, next); } };
tips：
一开始没理解好题意。
这个题要求必须走到某条path的叶子节点才算数，因此终止条件为走到叶子节点或者NULL。此外，root->left或者root->right不为NULL才往这个分支走。
原文：http://www.cnblogs.com/xbf9xbf/p/4508964.html