# 337. House-Robber-III

difficulty: Medium

&#x20;section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }&#x20;

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.
```

## Method One

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int rob(TreeNode root) {
        int[] res = dfs(root);
        return Math.max( res[0], res[1] );
    }
    public int[] dfs(TreeNode node) {
        int[] res = new int[2];
        if( node == null ) {
            return res;
        }

        int[] left = dfs(node.left);
        int[] right = dfs(node.right);

        res[0] = left[1] + right[1] + node.val; // Don't rob left or right, rob this 
        res[1] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]); // Doesn't have to rob children node.

        return res;
    }
}
​
/*
res[0] rob this one;
res[1] not rob this one;
*/
​
```
