LeetCode112 Path Sum

文章目录
  1. 1. 描述
  2. 2. 样例
  3. 3. 思路
  4. 4. 代码

描述

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.

样例

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Input:
Given the below binary tree and sum = 22
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
Output:
return true, as there exist a root-to-leaf
path 5->4->11->2 which sum is 22.

思路

给定一颗二叉树和一个值,问是否存在从根到叶的路径,满足路径权值和等于给定的值。

可以从根开始dfs,每经过一个节点就减去该结点的值,然后在叶结点判断给定的值是否被减到零。

代码

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/**
* 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;
sum -= root->val;
if (sum == 0 && root->left == NULL && root->right == NULL) return true;
return hasPathSum(root->left, sum) || hasPathSum(root->right, sum);
}
};
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