What is the primary technique to solve Time Taken to Mark All Nodes?

The primary technique is graph traversal using depth-first search (DFS), combined with dynamic programming to optimize the marking time calculation.

How do we optimize the DFS approach in this problem?

Dynamic programming is used to store intermediate results of the DFS traversal, avoiding redundant calculations and improving efficiency.

What should I focus on when solving the Time Taken to Mark All Nodes problem?

Focus on applying DFS for traversal and dynamic programming for time propagation, ensuring minimal computational overhead and handling large trees efficiently.

Why is dynamic programming useful in this problem?

Dynamic programming helps to optimize the solution by storing previously computed times for subtrees, preventing the need for recalculating them multiple times.

What is the expected time complexity for this problem?

The time complexity is generally O(n), as each node is visited once during the DFS traversal, and dynamic programming ensures efficient processing.

#3241

Hard

auto_awesome图·DFS·traversal

LeetCode 题解工作台

标记所有节点需要的时间

给你一棵无向树，树中节点从 0 到 n - 1 编号。同时给你一个长度为 n - 1 的二维整数数组 edges ，其中 edges[i] = [u i , v i ] 表示节点 u i 和 v i 在树中有一条边。一开始，所有节点都未标记。对于节点 i ：当 i 是奇数时，如果时刻…

动态规划树深度优先搜索图

题目描述

给你一棵无向树，树中节点从 0 到 n - 1 编号。同时给你一个长度为 n - 1 的二维整数数组 edges ，其中 edges[i] = [u_i, v_i] 表示节点 u_i 和 v_i 在树中有一条边。

一开始，所有节点都 未标记 。对于节点 i ：

当 i 是奇数时，如果时刻 x - 1 该节点有至少一个相邻节点已经被标记了，那么节点 i 会在时刻 x 被标记。
当 i 是偶数时，如果时刻 x - 2 该节点有至少一个相邻节点已经被标记了，那么节点 i 会在时刻 x 被标记。

请你返回一个数组 times ，表示如果你在时刻 t = 0 标记节点 i ，那么时刻 times[i] 时，树中所有节点都会被标记。

请注意，每个 times[i] 的答案都是独立的，即当你标记节点 i 时，所有其他节点都未标记。

示例 1：

输入：edges = [[0,1],[0,2]]

输出：[2,4,3]

解释：

对于 i = 0 ：
- 节点 1 在时刻 t = 1 被标记，节点 2 在时刻 t = 2 被标记。
对于 i = 1 ：
- 节点 0 在时刻 t = 2 被标记，节点 2 在时刻 t = 4 被标记。
对于 i = 2 ：
- 节点 0 在时刻 t = 2 被标记，节点 1 在时刻 t = 3 被标记。

示例 2：

输入：edges = [[0,1]]

输出：[1,2]

解释：

对于 i = 0 ：
- 节点 1 在时刻 t = 1 被标记。
对于 i = 1 ：
- 节点 0 在时刻 t = 2 被标记。

示例 3：

输入：edges = [[2,4],[0,1],[2,3],[0,2]]

输出：[4,6,3,5,5]

解释：

提示：

2 <= n <= 10⁵
edges.length == n - 1
edges[i].length == 2
0 <= edges[i][0], edges[i][1] <= n - 1
输入保证 edges 表示一棵合法的树。

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
Candidates should understand tree traversal techniques and be able to optimize with dynamic programming.
question_mark
Look for an approach that minimizes redundant calculations in tree traversal.
question_mark
Expect candidates to consider edge cases and tree sizes up to n = 10^5.

warning

常见陷阱

外企场景

error
Not optimizing the DFS traversal using dynamic programming, resulting in recalculating times for the same nodes.
error
Misunderstanding the propagation of times when transitioning from one subtree to another.
error
Overcomplicating the problem by attempting a brute-force solution rather than leveraging DFS and dynamic programming.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What happens if the tree is unbalanced? How does the approach scale?
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Can this approach be adapted to handle multiple starting nodes simultaneously?
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How would the solution change if nodes could be marked at times other than t=0?

help

常见问题

外企场景

继续练习

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