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
完