## [Leetcode]112. Path Sum(C++)

### 题目描述

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.

### 例子

Input: `[5,4,8,11,null,13,4,7,2,null,null,null,1]` Output: `22` Explanation: there exist a root-to-leaf path `5->4->11->2` which sum is 22.

### Note

• A leaf is a node with no children.

### 解题思路

• `root` 为空，返回 `false`
• `root` 为叶结点且其值等于 `sum` 返回 `true`
• 对其左右节点递归调用 `hasPathSum()`，将 `sum` 设为 `sum - root->val`，其中一棵子树返回 `true` 即返回 `true`

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
* right(right) {}
* };
*/
class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
if (!root) return false;
if (!root->left && !root->right && root->val == sum) return true;
return hasPathSum(root->left, sum - root->val) ||
hasPathSum(root->right, sum - root->val);
}
};
``````
• 时间复杂度: O(n)
• 空间复杂度: O(h) 递归深度

GitHub 代码同步地址： 112.PathSum.cpp

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