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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0301_0400.s0337_house_robber_iii;

// #Medium #Dynamic_Programming #Depth_First_Search #Tree #Binary_Tree #Udemy_Tree_Stack_Queue
// #2022_07_10_Time_1_ms_(91.77%)_Space_44.4_MB_(60.28%)

import com_github_leetcode.TreeNode;

/*
 * 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;
 *     }
 * }
 */
/**
 * 337 - House Robber III\.
 *
 * Medium
 *
 * The thief has found himself a new place for his thievery again. There is only one entrance to this area, called `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 form a binary tree. It will automatically contact the police if **two directly-linked houses were broken into on the same night**.
 *
 * Given the `root` of the binary tree, return _the maximum amount of money the thief can rob **without alerting the police**_.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/03/10/rob1-tree.jpg)
 *
 * **Input:** root = [3,2,3,null,3,null,1]
 *
 * **Output:** 7
 *
 * **Explanation:** Maximum amount of money the thief can rob = 3 + 3 + 1 = 7. 
 *
 * **Example 2:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/03/10/rob2-tree.jpg)
 *
 * **Input:** root = [3,4,5,1,3,null,1]
 *
 * **Output:** 9
 *
 * **Explanation:** Maximum amount of money the thief can rob = 4 + 5 = 9. 
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range [1, 10⁴].
 * *   0 <= Node.val <= 10⁴
**/
public class Solution {
    public int rob(TreeNode root) {
        int[] out = robRec(root);
        return Math.max(out[0], out[1]);
    }

    private int[] robRec(TreeNode curr) {
        if (curr == null) {
            return new int[] {0, 0};
        }
        int[] left = robRec(curr.left);
        int[] right = robRec(curr.right);
        int[] out = new int[2];
        // 1. If choosing to take the house
        out[0] = curr.val + left[1] + right[1];
        // 2. If choosing to skip the house
        out[1] = left[0] + right[0];
        // 3. Best Solution at house
        out[0] = Math.max(out[0], out[1]);
        return out;
    }
}