# 337. House Robber III

## Problem

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.

Example 1:

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

Example 2:

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

## Code

```
class Solution {
public int[] helper(TreeNode node) {
if (node == null) {
return new int[] { 0, 0 };
}
int left[] = helper(node.left);
int right[] = helper(node.right);
int rob = node.val + left[1] + right[1];
int notRob = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
return new int[] { rob, notRob };
}
public int rob(TreeNode root) {
int[] res = helper(root);
return Math.max(res[0], res[1]);
}
}
```

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338. Counting Bits