House Robbing

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.

Frankly I could not come up a smart solution to this at first sight, so that I took credit from https://www.hrwhisper.me/leetcode-house-robber-iii/
I really learnt a lot and got more understanding on DP after understanding it.

public class Solution {
    public int rob(TreeNode root) {
        return dfs(root)[0];
    }
   
    private int[] dfs(TreeNode root) {
        int dp[]={0,0};
        if(root != null){
            int[] dp_L = dfs(root.left);
            int[] dp_R = dfs(root.right);
            dp[1] = dp_L[0] + dp_R[0];
            dp[0] = Math.max(dp[1] ,dp_L[1] + dp_R[1] + root.val);
        }
        return dp;
    }
}