# Question

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.


# Solution

For each node, return the sum to rob it or not rob it. $$O(n)$$.

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

// [rob this, don't rob this]
private int[] robHelper(TreeNode root) {
int rob = root.val, noRob = 0;
if(root.left != null) {
int[] left = robHelper(root.left);
rob += left[1];
noRob += Math.max(left[0], left[1]);
}
if(root.right != null) {
int[] right = robHelper(root.right);
rob += right[1];
noRob += Math.max(right[0], right[1]);
}
return new int[]{rob, noRob};
}
}