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 hou…

小偷又发现一个新的可行窃的地点. 这个地区只有一个入口,称为“根”. 除了根部之外,每栋房子有且只有一个父房子. 一番侦察之后,聪明的小偷意识到“这个地方的所有房屋形成了一棵二叉树”. 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警.在不触动警报的情况下,计算小偷一晚能盗取的最高金额.示例 1:     3    / \   2   3    \   \      3   1能盗取的最高金额 = 3 + 3 + 1 = 7.示例 2:     3    / \   4   5  / \…

337. House Robber III Total Accepted: 18475 Total Submissions: 47725 Difficulty: Medium 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…

198. House Robber You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected…

House Robber:不能相邻,求能获得的最大值 House Robber II:不能相邻且第一个和最后一个不能同时取,求能获得的最大值 House Robber III:二叉树下的不能相邻,求能获得的最大值 Paint House:用3种颜色,相邻的房屋不能用同一种颜色,求花费最小 Paint House II:用k种颜色,相邻的房屋不能用同一种颜色,求花费最小Paint Fence:用k种颜色,相邻的可以用同一种颜色,但不能超过连续的2个,求有多少种可能性 198. House Robb…

337. 打家劫舍 III 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为"根". 除了"根"之外,每栋房子有且只有一个"父"房子与之相连.一番侦察之后,聪明的小偷意识到"这个地方的所有房屋的排列类似于一棵二叉树". 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警. 计算在不触动警报的情况下,小偷一晚能够盗取的最高金额. 示例 1: 输入: [3,2,3,nu…

337. 打家劫舍 III 题目链接 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/house-robber-iii 著作权归领扣网络所有.商业转载请联系官方授权,非商业转载请注明出处. 题目描述 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为"根". 除了"根"之外,每栋房子有且只有一个"父"房子与之相连.一番侦察之后,聪明的小偷…

Leetcode 337. 打家劫舍 III 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为"根". 除了"根"之外,每栋房子有且只有一个"父"房子与之相连.一番侦察之后,聪明的小偷意识到"这个地方的所有房屋的排列类似于一棵二叉树". 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警. 计算在不触动警报的情况下,小偷一晚能够盗取的最高金额. 输入: [3,2,3…

198. House Robber You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected…

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 hou…

题目描述 小偷又发现一个新的可行窃的地点. 这个地区只有一个入口,称为“根”. 除了根部之外,每栋房子有且只有一个父房子. 一番侦察之后,聪明的小偷意识到“这个地方的所有房屋形成了一棵二叉树”. 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警. 在不触动警报的情况下,计算小偷一晚能盗取的最高金额. 示例 1: / \ 2 3 \ \ 3 1 能盗取的最高金额 = + + = 7. 示例 2: 3 / \ / \ \ 1 3 1 能盗取的最高金额 = + = 9. 解题思路 用后序遍历的…

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 hou…

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 hou…

题目描述: 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 "a…

这一篇也是基于"打家劫舍"的扩展,需要针对特殊情况特殊考虑,当然其本质还是动态规划,优化时需要考虑数据结构. 原题 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为"根". 除了"根"之外,每栋房子有且只有一个"父"房子与之相连.一番侦察之后,聪明的小偷意识到"这个地方的所有房屋的排列类似于一棵二叉树". 如果两个直接相连的房子在同一天晚上被打劫,房屋…

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 hou…

二刷吧..不知道为什么house robbery系列我找不到笔记,不过印象中做了好几次了. 不是很难,用的post-order做bottom-up的运算. 对于一个Node来说,有2种情况,一种是选(自己+下下层):一种是选左右children. 其实就是选自己和不选自己的区别.其实更像是dfs的题而不是DP的题. Time: O(n) Space: O(lgn) public class Solution { public int rob(TreeNode root) { if (root =…

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 hou…

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 hou…

［抄题］: 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 "a…

题目链接 https://leetcode.com/problems/house-robber-iii/description/ 题目描述 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为"根". 除了"根"之外,每栋房子有且只有一个"父"房子与之相连.一番侦察之后,聪明的小偷意识到"这个地方的所有房屋的排列类似于一棵二叉树". 如果两个直接相连的房子在同一天晚上被打劫,…

Total Accepted: 1341 Total Submissions: 3744 Difficulty: Medium 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…

一.题目描述 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为“根”. 除了“根”之外,每栋房子有且只有一个“父“房子与之相连.一番侦察之后,聪明的小偷意识到“这个地方的所有房屋的排列类似于一棵二叉树”. 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警. 计算在不触动警报的情况下,小偷一晚能够盗取的最高金额. 示例 1: 输入: [3,2,3,null,3,null,1] 3 / \ 2 3 \ \ 3 1 输出: 7 解释: 小…

题目描述: 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为“根”. 除了“根”之外,每栋房子有且只有一个“父“房子与之相连.一番侦察之后,聪明的小偷意识到“这个地方的所有房屋的排列类似于一棵二叉树”. 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警. 计算在不触动警报的情况下,小偷一晚能够盗取的最高金额. 示例 1: 输入: [3,2,3,null,3,null,1] 3 / \ 2 3 \ \ 3 1 输出: 7 解释: 小偷…

题目描述: 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为“根”. 除了“根”之外,每栋房子有且只有一个“父“房子与之相连.一番侦察之后,聪明的小偷意识到“这个地方的所有房屋的排列类似于一棵二叉树”. 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警. 计算在不触动警报的情况下,小偷一晚能够盗取的最高金额. 示例 1: 输入: [3,2,3,null,3,null,1] 3 / \ 2 3 \ \ 3 1 输出: 7 解释: 小偷…

每个节点是个房间,数值代表钱.小偷偷里面的钱,不能偷连续的房间,至少要隔一个.问最多能偷多少钱 TreeNode* cur mp[{cur, true}]表示以cur为根的树,最多能偷的钱 mp[{cur, false}]表示以cur为根的树,不偷cur节点的钱,最多能偷的钱 可以看出有下面的关系 mp[{node, false}] = mp[{node->left,true}] + mp[{node->right,true}] mp[{node, true}] = max(node->…

题目 在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为"根". 除了"根"之外,每栋房子有且只有一个"父"房子与之相连.一番侦察之后,聪明的小偷意识到"这个地方的所有房屋的排列类似于一棵二叉树". 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警. 计算在不触动警报的情况下,小偷一晚能够盗取的最高金额. 示例 1: 输入: [3,2,3,null,3,null,1…

在上次打劫完一条街道之后和一圈房屋后,小偷又发现了一个新的可行窃的地区.这个地区只有一个入口,我们称之为"根". 除了"根"之外,每栋房子有且只有一个"父"房子与之相连.一番侦察之后,聪明的小偷意识到"这个地方的所有房屋的排列类似于一棵二叉树". 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警.计算在不触动警报的情况下,小偷一晚能够盗取的最高金额.示例 1:输入: [3,2,3,null,3,null,1]    …

小偷又发现一个新的可行窃的地点. 这个地区只有一个入口,称为"根". 除了根部之外,每栋房子有且只有一个父房子. 一番侦察之后,聪明的小偷意识到"这个地方的所有房屋形成了一棵二叉树". 如果两个直接相连的房子在同一天晚上被打劫,房屋将自动报警. 在不触动警报的情况下,计算小偷一晚能盗取的最高金额. 示例 1: 3 / \ 2 3 \ \ 3 1 能盗取的最高金额 = 3 + 3 + 1 = 7. 示例 2: 3 / \ 4 5 / \ \ 1 3 1 能盗取的最高金…

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…