/** * 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==NULL) return false; if(root->left ==NULL && root->right == NULL) return root->val==sum; if(hasPathSum(root->left,sum-root->val))return true; if(hasPathSum(root->right,sum-root->val))return true; return false; } };
原文:http://www.cnblogs.com/julie-yang/p/4668768.html