# 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:
3
/ \
2   3
\   \
3   1
Maximum amount of money the thief can rob = 3 + 3 + 1 = 7.
Example 2:
3
/ \
4   5
/ \   \
1   3   1
Maximum amount of money the thief can rob = 4 + 5 = 9.
```

### Solutions:

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private class Info {
public int rob;
public int norob;
public Info(int rob, int norob) {
this.rob = rob;
this.norob = norob;
}
}
public int rob(TreeNode root) {
Info info = dfs(root);
return Math.max(info.rob, info.norob);
}
private Info dfs(TreeNode node) {
if (node == null) {
return new Info(0, 0);
}
Info left = dfs(node.left);
Info right = dfs(node.right);
Info result = new Info(left.norob + right.norob + node.val, Math.max(left.rob, left.norob) + Math.max(right.rob, right.norob));
return result;
}
}
``````