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.
解体思路：深度优先遍历，使用递归。
/** * 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 == nullptr) return false; if(root->left==nullptr && root->right == nullptr) return root->val == sum; return hasPathSum(root->left, sum - root->val)|| hasPathSum(root->right, sum - root->val); } };
原文：http://www.cnblogs.com/zxy1992/p/4386829.html