What is the time complexity of Minimum Height Trees?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Height Trees?

This problem uses Depth-First Search, Breadth-First Search, Graph, Topological Sort. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Height Trees?

Insight 1: The MHTs are essentially the middle nodes of the longest path in the tree. If the longest path has an odd number of nodes, there is one middle node. If it has an even number of nodes, there are two middle nodes. Insight 2: Using a topological sort approach, we can iteratively remove leaf nodes (nodes with degree 1) until only the middle node(s) remain. These remaining nodes will be the roots of the MHTs.

Which companies ask LeetCode 310. Minimum Height Trees in interviews?

This problem is frequently asked at Citadel, PhonePe, Splunk, Stackline.

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Minimum Height Trees - LeetCode 310 Solution

Minimum Height Trees - Complete Solution Guide

Minimum Height Trees is LeetCode problem 310, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree. Given a tree of n nodes labelled from 0 to n - 1 , and an array of n - 1 edges where edges[i] = [a i , b i ] indicates that there is an undirected edge between the two nodes a i and b i in the tree, you can choose any node of the tree as the root. When you select a node x as the root, the result tree has height h . Among all possible root

Detailed Explanation

The problem asks us to find the minimum height trees (MHTs) of a given undirected tree. We are given the number of nodes `n` and a list of edges `edges` representing the tree. We need to return a list of the root labels of all possible MHTs. The height of a rooted tree is defined as the number of edges on the longest path from the root to a leaf. An MHT is a tree with the minimum possible height among all possible rooted trees formed by choosing different nodes as the root.

Solution Approach

The provided solution uses a topological sort-like approach, akin to a breadth-first search (BFS), to iteratively remove leaf nodes. It starts by identifying all leaf nodes in the tree and placing them in a queue. Then, it repeatedly removes leaf nodes from the queue and updates the degrees of their neighbors. If a neighbor's degree becomes 1 after removing a leaf, it is added to the queue. This process continues until only two or fewer nodes remain. These remaining nodes are the roots of the MHTs.

Step-by-Step Algorithm

Step 1: Handle the base case where n = 1. In this case, the only node (0) is the root of the MHT.
Step 2: Build an adjacency list (`adj`) to represent the graph, and an array `degrees` to store the degree of each node.
Step 3: Initialize a queue `leaves` with all nodes that have a degree of 1 (leaf nodes).
Step 4: While the number of remaining nodes (`remaining_nodes`) is greater than 2:
a. Remove all current leaf nodes from the `leaves` queue.
b. For each removed leaf node, decrement the degree of its neighbors.
c. If a neighbor's degree becomes 1 after decrementing, add it to the `leaves` queue.
d. Update `remaining_nodes` by subtracting the number of removed leaves.
Step 5: The remaining nodes in the `leaves` queue are the roots of the MHTs. Return them as a list.

Key Insights

Insight 1: The MHTs are essentially the middle nodes of the longest path in the tree. If the longest path has an odd number of nodes, there is one middle node. If it has an even number of nodes, there are two middle nodes.
Insight 2: Using a topological sort approach, we can iteratively remove leaf nodes (nodes with degree 1) until only the middle node(s) remain. These remaining nodes will be the roots of the MHTs.
Insight 3: The algorithm simulates pruning the tree from the leaves, layer by layer, until we reach the center(s). This is analogous to peeling an onion; the last layer(s) left are the core(s) we want.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Depth-First Search, Breadth-First Search, Graph, Topological Sort.

Companies

Asked at: Citadel, PhonePe, Splunk, Stackline.

Minimum Height Trees | LeetCode 310

Medium

Depth-First Search Breadth-First Search Graph Topological Sort

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        if (n == 1) {
            return Collections.singletonList(0);
        }

        Map<Integer, List<Integer>> adj = new HashMap<>();
        int[] degrees = new int[n];
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.computeIfAbsent(u, k -> new ArrayList<>()).add(v);
            adj.computeIfAbsent(v, k -> new ArrayList<>()).add(u);
            degrees[u]++;
            degrees[v]++;
        }

        Queue<Integer> leaves = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if (degrees[i] == 1) {
                leaves.offer(i);
            }
        }

        int remaining_nodes = n;
        while (remaining_nodes > 2) {
            int num_leaves = leaves.size();
            remaining_nodes -= num_leaves;
            for (int i = 0; i < num_leaves; i++) {
                int leaf = leaves.poll();
                for (int neighbor : adj.get(leaf)) {
                    degrees[neighbor]--;
                    if (degrees[neighbor] == 1) {
                        leaves.offer(neighbor);
                    }
                }
            }
        }

        return new ArrayList<>(leaves);
    }
}

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Problem Info

Description

Problem 310: Minimum Height Trees - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Graph Topological Sort

Companies

CitadelPhonePeSplunkStackline

Stats

Difficulty:Medium

Languages:4

Companies:4

Problem #:310

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Handle the base case where n = 1. In this case, the only node (0) is the root of the MHT.
2
Step 2: Build an adjacency list (`adj`) to represent the graph, and an array `degrees` to store the degree of each node.
3
Step 3: Initialize a queue `leaves` with all nodes that have a degree of 1 (leaf nodes).
4
Step 4: While the number of remaining nodes (`remaining_nodes`) is greater than 2:
5
a. Remove all current leaf nodes from the `leaves` queue.
6
b. For each removed leaf node, decrement the degree of its neighbors.
7
c. If a neighbor's degree becomes 1 after decrementing, add it to the `leaves` queue.
8
d. Update `remaining_nodes` by subtracting the number of removed leaves.
9
Step 5: The remaining nodes in the `leaves` queue are the roots of the MHTs. Return them as a list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the need to traverse each edge a limited number of times. The space complexity is derived from storing the adjacency list and degrees, which scales linearly with the number of nodes.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Depth-First Search (DFS)

You can theoretically use DFS to calculate the height for each node when it's set as root, and then find the node with the minimum height. However, this approach will be significantly slower.

Time: O(n^2)Space: O(n)

Trade-offs: This approach is less efficient due to the repeated calculation of tree heights for each potential root. The provided solution is much faster.

✨

Key Insights

1Insight 1: The MHTs are essentially the middle nodes of the longest path in the tree. If the longest path has an odd number of nodes, there is one middle node. If it has an even number of nodes, there are two middle nodes.
2Insight 2: Using a topological sort approach, we can iteratively remove leaf nodes (nodes with degree 1) until only the middle node(s) remain. These remaining nodes will be the roots of the MHTs.
3Insight 3: The algorithm simulates pruning the tree from the leaves, layer by layer, until we reach the center(s). This is analogous to peeling an onion; the last layer(s) left are the core(s) we want.

📚

Key Terms Explained

For Students

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: n = 1. The code handles this by returning `[0]` directly, which is correct.
Edge case 2: A star graph where one node is connected to all other nodes. The algorithm will correctly identify the center node after removing all leaf nodes.
Edge case 3: A perfectly balanced binary tree. The algorithm will correctly identify the root node.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the base case of n = 1 correctly. This can lead to errors or incorrect results.

✗Common mistake 2: Failing to update the degrees of neighbors when removing a leaf node. This can cause the algorithm to incorrectly identify leaf nodes in subsequent iterations.

✗Common mistake 3: Not understanding the core concept that the MHT roots are located near the center of the longest paths in the tree and attempting to use brute-force height calculations, which are significantly less efficient.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network routing. Finding the optimal server location in a network to minimize latency for all clients. The MHT approach can help determine the most central node in the network graph.

🏦Real-world scenario 2: File system synchronization. When synchronizing files across multiple devices, choosing the most central node as the synchronization point can minimize the overall synchronization time.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Height Trees"?

For this medium-level Depth-First Search problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?