Middle-题目129：337. House Robber III（增补1）

最新推荐文章于 2019-07-06 10:27:46 发布

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解决一个贼如何在二叉树结构的社区中偷窃而不被发现的问题，通过递归算法实现，确保相邻房屋不会在同一晚被盗。

题目原文：
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.
题目大意：
此题是Easy-题目26和Middle-题目58的变形，这次这个贼偷的是二叉树，每偷一个节点，它的父节点和子节点都不能偷了，求最大收益。
题目分析：
我们很熟悉的前两题都是使用动态规划，而这道题用的是深搜，深搜的返回结果是一个两个元素的数组profit，其中profit[0]代表偷当前节点的时候得到的最大收益，profit[1]代表不偷当前节点时获得的最大收益。那么每次搜索左右子树的最大收益记为leftprofit和rightprofit，并取：
$profit[0] = leftProfit[1] + rightProfit[1] + root.val;$
$profit[1] = Math.max(leftProfit[0],leftProfit[1]) + Math.max(rightProfit[0], rightProfit[1]);$
直到遇到空节点返回new int[2].
源码：（language：java）

public class Solution {
    public int rob(TreeNode root) {
        int[] profit = dfs(root);
        return Math.max(profit[0], profit[1]);
    }
    private int[] dfs(TreeNode root) {
        int[] profit = new int[2];
        // profit[0] stands for the maximum profit with robbing root, while profit[1] stands for the maximum profit without robbing root.
        if(root!=null) {
            int[] leftProfit = dfs(root.left);
            int[] rightProfit = dfs(root.right);
            profit[0] = leftProfit[1] + rightProfit[1] + root.val;
            profit[1] = Math.max(leftProfit[0],leftProfit[1]) + Math.max(rightProfit[0], rightProfit[1]);
            return profit;
        }
        return new int[2];
    }
}