What is the time complexity of Maximum Width of Binary Tree?

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

What data structures are used to solve Maximum Width of Binary Tree?

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

What are the key insights for solving Maximum Width of Binary Tree?

Insight 1: We can use Breadth-First Search (BFS) to traverse the tree level by level. Insight 2: We need to assign a position to each node based on its potential location in a complete binary tree. The root node has position 0. If a node has position 'p', its left child has position '2*p' and its right child has position '2*p + 1'.

Which companies ask LeetCode 662. Maximum Width of Binary Tree in interviews?

This problem is frequently asked at Flipkart.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Maximum Width of Binary Tree - LeetCode 662 Solution

Maximum Width of Binary Tree - Complete Solution Guide

Maximum Width of Binary Tree is LeetCode problem 662, 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

Given the root of a binary tree, return the maximum width of the given tree . The maximum width of a tree is the maximum width among all levels. The width of one level is defined as the length between the end-nodes (the leftmost and rightmost non-null nodes), where the null nodes between the end-nodes that would be present in a complete binary tree extending down to that level are also counted into the length calculation. It is guaranteed that the answer will in the range of a 32-bit signed inte

Detailed Explanation

The problem asks us to find the maximum width of a given binary tree. The width of a level is defined as the number of nodes between the leftmost and rightmost non-null nodes, inclusive, considering the binary tree as a complete binary tree. Essentially, we need to calculate the distance between the leftmost and rightmost node at each level and return the maximum of these distances. The nodes are implicitly assigned positions based on their position in a complete binary tree, starting with position 0 for the root.

Solution Approach

The solution uses a Breadth-First Search (BFS) algorithm to traverse the binary tree level by level. For each node, it stores the node and its position within a complete binary tree. As BFS explores each level, it calculates the width of the level by subtracting the position of the leftmost node from the position of the rightmost node and adding 1. It keeps track of the maximum width encountered so far and returns it at the end.

Step-by-Step Algorithm

Step 1: Initialize a queue with the root node and its position (0). Initialize the maximum width to 0.
Step 2: While the queue is not empty, get the number of nodes at the current level.
Step 3: Determine the leftmost and rightmost positions on the current level.
Step 4: Calculate the width of the current level: `rightmost_pos - leftmost_pos + 1`.
Step 5: Update the maximum width if the current level's width is greater.
Step 6: Iterate through all the nodes at the current level. For each node, dequeue it, get its position, and calculate the positions of its left and right children (if they exist). Enqueue the left child with position `2 * (current_node_position - leftmost_position)` and the right child with position `2 * (current_node_position - leftmost_position) + 1`.
Step 7: Repeat steps 2-6 until the queue is empty.
Step 8: Return the maximum width.

Key Insights

Insight 1: We can use Breadth-First Search (BFS) to traverse the tree level by level.
Insight 2: We need to assign a position to each node based on its potential location in a complete binary tree. The root node has position 0. If a node has position 'p', its left child has position '2*p' and its right child has position '2*p + 1'.
Insight 3: To avoid integer overflow, it's important to normalize the positions at each level by subtracting the leftmost position of that level from all positions on that level. This ensures that the position values don't grow too large.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Tree, Depth-First Search, Breadth-First Search, Binary Tree.

Companies

Asked at: Flipkart.

Maximum Width of Binary Tree | LeetCode 662

Medium

Tree Depth-First Search Breadth-First Search Binary Tree

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

Solution Code

Available in:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
import java.util.Deque;
import java.util.LinkedList;

class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        if (root == null) {
            return 0;
        }

        int max_width = 0;
        Deque<Pair<TreeNode, Integer>> queue = new LinkedList<>();
        queue.offer(new Pair<>(root, 0));

        while (!queue.isEmpty()) {
            int level_length = queue.size();
            int level_start_pos = queue.peekFirst().getValue();

            int rightmost_pos = 0;
            for (Pair<TreeNode, Integer> p : queue) {
                rightmost_pos = p.getValue();
            }
            

            max_width = Math.max(max_width, rightmost_pos - level_start_pos + 1);

            for (int i = 0; i < level_length; i++) {
                Pair<TreeNode, Integer> pair = queue.pollFirst();
                TreeNode node = pair.getKey();
                int pos = pair.getValue();

                int relative_pos = pos - level_start_pos;

                if (node.left != null) {
                    queue.offer(new Pair<>(node.left, 2 * relative_pos));
                }
                if (node.right != null) {
                    queue.offer(new Pair<>(node.right, 2 * relative_pos + 1));
                }
            }
        }

        return max_width;
    }

    static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 662: Maximum Width of Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

Flipkart

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:662

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

8 steps

1
Step 1: Initialize a queue with the root node and its position (0). Initialize the maximum width to 0.
2
Step 2: While the queue is not empty, get the number of nodes at the current level.
3
Step 3: Determine the leftmost and rightmost positions on the current level.
4
Step 4: Calculate the width of the current level: `rightmost_pos - leftmost_pos + 1`.
5
Step 5: Update the maximum width if the current level's width is greater.
6
Step 6: Iterate through all the nodes at the current level. For each node, dequeue it, get its position, and calculate the positions of its left and right children (if they exist). Enqueue the left child with position `2 * (current_node_position - leftmost_position)` and the right child with position `2 * (current_node_position - leftmost_position) + 1`.
7
Step 7: Repeat steps 2-6 until the queue is empty.
8
Step 8: Return the maximum width.

💡

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 O(N) because each node is visited and processed exactly once. The space complexity is determined by the maximum number of nodes that can be present in the queue at any given time during the BFS traversal, which corresponds to the width of the widest level.

📖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

Depth-First Search (DFS)

It is possible to solve this problem using DFS, but it is less intuitive. We would need to maintain a list of leftmost positions for each level and update the maximum width during the DFS traversal.

Time: O(N)Space: O(H), where H is the height of the tree. In the worst case (skewed tree), H can be N, so the space complexity is O(N).

Trade-offs: BFS is generally preferred due to its iterative nature and easier level-by-level processing. DFS can be more complex to implement for this specific problem.

✨

Key Insights

1Insight 1: We can use Breadth-First Search (BFS) to traverse the tree level by level.
2Insight 2: We need to assign a position to each node based on its potential location in a complete binary tree. The root node has position 0. If a node has position 'p', its left child has position '2*p' and its right child has position '2*p + 1'.
3Insight 3: To avoid integer overflow, it's important to normalize the positions at each level by subtracting the leftmost position of that level from all positions on that level. This ensures that the position values don't grow too large.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (root is null). The solution correctly returns 0 in this case.
Edge case 2: Tree with only one node (root only). The solution returns 1 in this case.
Edge case 3: Skewed tree (either left or right). The solution correctly handles the position assignments and calculates the width.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not normalizing the positions at each level. This can lead to integer overflow, especially for deeper trees.

✗Common mistake 2: Incorrectly calculating the positions of child nodes. Ensure the positions are calculated as `2 * parent_position` for the left child and `2 * parent_position + 1` for the right child.

✗Common mistake 3: Forgetting to add 1 when calculating the width: `rightmost_pos - leftmost_pos + 1`.

✅

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: Representing organizational charts where calculating the width of each level can determine the span of control.

🏦Real-world scenario 2: Analyzing network topologies represented as trees where the width can indicate the degree of parallelism at a particular layer.

🎮Real-world scenario 3: File system representation. While a true file system is a graph not a tree, tree-like sub-structures can be analyzed for depth and 'width' of directories.

💡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

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

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 1, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

❓

Frequently Asked Questions

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

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 "Maximum Width of Binary Tree"?

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

🤖 Need help with this problem?