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.
基本思路:
进行深度递归。
逐层从sum中扣去当前val值;到叶结点时,看剩下的sum和val值是否相等。
/** * 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) return false; if (!root->left && !root->right && sum == root->val) return true; return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val); } };
时间: 2024-10-13 16:55:09