What is the main approach for solving the Minimum Weighted Subgraph With the Required Paths II problem?

The problem can be solved using binary-tree traversal combined with techniques like binary lifting to optimize path queries and state tracking for efficient subtree weight calculation.

How does binary lifting help in solving the Minimum Weighted Subgraph With the Required Paths II?

Binary lifting optimizes the process of finding the Lowest Common Ancestor (LCA) of nodes, which is crucial for minimizing the path weights between source and destination nodes in the tree.

What is the time complexity of the best solution for the Minimum Weighted Subgraph With the Required Paths II?

The optimal time complexity is O(n log n) for preprocessing with O(log n) per query, depending on the approach used for tree traversal and binary lifting.

What are some common mistakes when solving the Minimum Weighted Subgraph With the Required Paths II?

Common mistakes include not optimizing the tree traversal, overlooking the importance of minimizing subtree weights, and failing to apply binary lifting correctly.

How can GhostInterview assist with solving the Minimum Weighted Subgraph With the Required Paths II problem?

GhostInterview provides hints and solutions for implementing efficient tree traversal, binary lifting, and state tracking techniques, which are essential for solving this problem efficiently.

#3553

Hard

auto_awesome二分·树·traversal

LeetCode 题解工作台

包含要求路径的最小带权子图 II

给你一个无向带权树，共有 n 个节点，编号从 0 到 n - 1 。这棵树由一个二维整数数组 edges 表示，长度为 n - 1 ，其中 edges[i] = [u i , v i , w i ] 表示存在一条连接节点 u i 和 v i 的边，权重为 w i 。此外，给你一个二维整数数组 …

数组树深度优先搜索

题目描述

给你一个 无向带权 树，共有 n 个节点，编号从 0 到 n - 1。这棵树由一个二维整数数组 edges 表示，长度为 n - 1，其中 edges[i] = [u_i, v_i, w_i] 表示存在一条连接节点 u_i 和 v_i 的边，权重为 w_i。

此外，给你一个二维整数数组 queries，其中 queries[j] = [src1_j, src2_j, dest_j]。

返回一个长度等于 queries.length 的数组 answer，其中 answer[j] 表示一个子树的 最小总权重 ，使用该子树的边可以从 src1_j 和 src2_j 到达 dest_j。

这里的子树是指原树中任意节点和边组成的连通子集形成的一棵有效树。

示例 1：

输入： edges = [[0,1,2],[1,2,3],[1,3,5],[1,4,4],[2,5,6]], queries = [[2,3,4],[0,2,5]]

输出： [12,11]

解释：

蓝色边表示可以得到最优答案的子树之一。

answer[0]：在选出的子树中，从 src1 = 2 和 src2 = 3 到 dest = 4 的路径总权重为 3 + 5 + 4 = 12。
answer[1]：在选出的子树中，从 src1 = 0 和 src2 = 2 到 dest = 5 的路径总权重为 2 + 3 + 6 = 11。

示例 2：

输入： edges = [[1,0,8],[0,2,7]], queries = [[0,1,2]]

输出： [15]

解释：

answer[0]：选出的子树中，从 src1 = 0 和 src2 = 1 到 dest = 2 的路径总权重为 8 + 7 = 15。

提示：

3 <= n <= 10⁵
edges.length == n - 1
edges[i].length == 3
0 <= u_i, v_i < n
1 <= w_i <= 10⁴
1 <= queries.length <= 10⁵
queries[j].length == 3
0 <= src1_j, src2_j, dest_j < n
src1_j、src2_j 和 dest_j 互不不同。
输入数据保证 edges 表示的是一棵有效的树。

lightbulb

解题思路

方法一

1

2

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Understanding how to apply DFS/BFS with binary lifting is key to solving this problem efficiently.
question_mark
The candidate should demonstrate awareness of tree traversal and path optimization strategies.
question_mark
Look for candidates who focus on time complexity optimization when handling large input sizes.

warning

常见陷阱

外企场景

error
Not efficiently handling large trees and queries, leading to timeouts or inefficient solutions.
error
Overlooking the need to minimize the weight of the subtree for each query, which can lead to incorrect answers.
error
Failing to apply binary lifting correctly, causing unnecessary recalculations of paths.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Optimizing the algorithm for a higher number of nodes and queries.
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Extending the problem to support dynamic updates to the tree or queries.
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Handling multiple types of tree structures beyond binary trees, such as general graphs.

help

常见问题

外企场景

继续练习

#3559 给边赋权值的方案数 II #3544 子树反转和 #3562 折扣价交易股票的最大利润