Is Find Diameter Endpoints of a Tree easy or hard?

The problem is generally classified as Medium difficulty. The core challenge is recognizing the tree diameter property that allows the double BFS optimization. Once that insight is known, the implementation is straightforward graph traversal.

Find Diameter Endpoints of a Tree Python/Java solution

Most implementations build an adjacency list and run two BFS traversals. The first BFS finds the farthest node from an arbitrary start, and the second BFS finds the farthest node from that result. This pattern is identical across Python, Java, C++, Go, TypeScript, and Rust.

How to solve Find Diameter Endpoints of a Tree in O(n)?

Use the double BFS diameter technique. First perform BFS from any node to find the farthest node A. Then run BFS again starting from A to find the farthest node B. Because the input is a tree with no cycles and exactly n−1 edges, both traversals together take O(n) time.

What is the best approach for Find Diameter Endpoints of a Tree?

The most efficient approach is the two-pass BFS technique. Run a BFS from any node to find the farthest node A, then run another BFS from A to find the farthest node B. The pair (A, B) forms the diameter endpoints. This method runs in O(n) time and O(n) space for a tree with n nodes.

Is Find Diameter Endpoints of a Tree asked at Google/Amazon/Meta?

Tree diameter problems frequently appear in interviews at companies like Google, Amazon, and Meta. Variants include computing the diameter length, returning the path, or identifying endpoints. Understanding the two-pass BFS or DFS technique is commonly expected in graph and tree interview rounds.

What data structure is used in Find Diameter Endpoints of a Tree?

The problem models the tree as an adjacency list graph. A queue is used for Breadth-First Search to traverse nodes level by level and compute distances. Arrays or hash maps typically track visited nodes and distances during traversal.

What is the time complexity of Find Diameter Endpoints of a Tree?

The optimal solution runs in O(n) time because each BFS traversal visits every node and edge once. Since the algorithm performs two BFS passes, the total work is still linear. Space complexity is O(n) for the adjacency list and queue used during traversal.

3787. Find Diameter Endpoints of a Tree - Practice Online

You are given an undirected tree with n nodes, numbered from 0 to n - 1. It is represented by a 2D integer array edges of length n - 1, where edges[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree.

A node is called special if it is an endpoint of any diameter path of the tree.

Return a binary string s of length n, where s[i] = '1' if node i is special, and s[i] = '0' otherwise.

A diameter path of a tree is the longest simple path between any two nodes. A tree may have multiple diameter paths.

An endpoint of a path is the first or last node on that path.

Example 1:

Input: n = 3, edges = [[0,1],[1,2]]

Output: "101"

Explanation:

The diameter of this tree consists of 2 edges.
The only diameter path is the path from node 0 to node 2
The endpoints of this path are nodes 0 and 2, so they are special.

Example 2:

Input: n = 7, edges = [[0,1],[1,2],[2,3],[3,4],[3,5],[1,6]]

Output: "1000111"

Explanation:

The diameter of this tree consists of 4 edges. There are 4 diameter paths:

The path from node 0 to node 4
The path from node 0 to node 5
The path from node 6 to node 4
The path from node 6 to node 5

The special nodes are nodes 0, 4, 5, 6, as they are endpoints in at least one diameter path.

Example 3:

Input: n = 2, edges = [[0,1]]

Output: "11"

Explanation:

The diameter of this tree consists of 1 edge.
The only diameter path is the path from node 0 to node 1
The endpoints of this path are nodes 0 and 1, so they are special.

Constraints:

2 <= n <= 10⁵
edges.length == n - 1
edges[i] = [a_i, b_i]
0 <= a_i, b_i < n
The input is generated such that edges represents a valid tree.