题目

balanced-binary-tree

2. 算法

* dfs

3. 代码

class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if (root == NULL) return false;
        if (root->left == NULL && root->right == NULL && root->val == sum ) return true;
        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }
};