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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: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：DFS。只要找到叶节点的时候，sum==0，就可以返回true；否则返回false。
代码：
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) return false; 14 if(!root->left&&!root->rigth) return root->val==sum;//判断是否为叶节点 15 return hasPathSum(root->left,root->val-sum)||hasPathSum(root->right,root->val-sum); 16 } 17 };
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原文地址：http://www.cnblogs.com/Deribs4/p/5686922.html
