LeetCode 337. House Robber III（小偷游戏）

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本文探讨了LeetCode上的House Robber III问题，通过树形结构动态规划的方法，设计了一个递归算法，该算法能够计算出在不触动警报的前提下，窃贼能获取的最大金额。

原题网址：https://leetcode.com/problems/house-robber-iii/

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:

Maximum amount of money the thief can rob = 3 + 3 + 1 = 7 .

Example 2:

Maximum amount of money the thief can rob = 4 + 5 = 9 .

方法：类似动态规划，递归遍历所有节点，返回两种结果，一种是包含本节点的最大值，另一种是不含本节点的最大值。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int[] dp(TreeNode root) {
        // money[0] root inclusive, money[1] root excluded
        int[] money = new int[2];
        money[0] = root.val;
        if (root.left != null) {
            int[] left = dp(root.left);
            money[0] += left[1];
            money[1] += Math.max(left[0], left[1]);
        }
        if (root.right != null) {
            int[] right = dp(root.right);
            money[0] += right[1];
            money[1] += Math.max(right[0], right[1]);
        }
        return money;
    }
    public int rob(TreeNode root) {
        if (root == null) return 0;
        int[] money = dp(root);
        return Math.max(money[0], money[1]);
    }
}

另一种实现：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int[] dp(TreeNode root) {
        // m[0] 最大收益（无论偷还是不偷本房屋）
        // m[1] 最大收益（不偷本房屋）
        int[] m = new int[2];
        m[0] = root.val;
        if (root.left != null) {
            int[] left = dp(root.left);
            m[1] += left[0];
            m[0] += left[1];
        }
        if (root.right != null) {
            int[] right = dp(root.right);
            m[1] += right[0];
            m[0] += right[1];
        }
        m[0] = Math.max(m[0], m[1]);
        return m;
    }
    public int rob(TreeNode root) {
        if (root == null) return 0;
        int[] m = dp(root);
        return m[0];
    }
}