0%

### 题目描述

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the “root.” Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that “all houses in this place forms a binary tree”. It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

### 例子

#### 例子 1

Input: `[3,2,3,null,3,null,1]`
Output: 7
Explanation: Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.

#### 例子 2

Input: `[3,4,5,1,3,null,1]`
Output: 9
Explanation: Maximum amount of money the thief can rob = 4 + 5 = 9.

### 解题思路

• 使用相隔一层之后的结果加上当前节点的值
• 直接使用相邻层的结果

`rob(root) = max(root->val + rob(two childrens of (root->left)) + rob(two children of root->right), rob(root->left) + rob(root->right))`

• 时间复杂度: O(n)
• 空间复杂度: O(n)

GitHub 代码同步地址： 337.HouseRobberIII.cpp

GitHub: Leetcode-C++-Solution