problem

 code

回归方法

/** * 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->val==sum)&&(root->left==NULL)&&(root->right==NULL) ) return true;//        return (hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val));            }};

 

 

 

 参考

1.;

 完