LeetCode 刷题笔记 之 House Robber III

xiaoxiao2021-02-28  3

题目如下:

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.

解答如下:

本来想简单地用递归去查找奇数层的和以及偶数层的和。但是忽略了。跨两层的和可能更大,比如:7----2----1----4。所以这题的正解还是动态规划。

class Solution { public int rob(TreeNode root) { return dfs(root)[1]; } private int[] dfs(TreeNode root) { int[] rob ={0, 0}; if(root != null) { int[] robLeft = dfs(root.left); int[] robRight = dfs(root.right); rob[0] = robLeft[1] + robRight[1]; //左子树的最大值,右子树的最大值 rob[1] = Math.max(robLeft[0] + robRight[0] + root.val, rob[0]); //要么是子树层,要么是本层+子树的子树层。 } return rob; } }
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