What is the time complexity of Diameter of Binary Tree?

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

What data structures are used to solve Diameter of Binary Tree?

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

What are the key insights for solving Diameter of Binary Tree?

Insight 1: The diameter of a tree at a given node can be computed by summing the depths of its left and right subtrees. Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the depth of each subtree. While calculating the depth, we can simultaneously update the diameter.

Which companies ask LeetCode 543. Diameter of Binary Tree in interviews?

This problem is frequently asked at Aurora, LinkedIn, Verkada, Visa, Wix.

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Diameter of Binary Tree - LeetCode 543 Solution

Diameter of Binary Tree - Complete Solution Guide

Diameter of Binary Tree is LeetCode problem 543, a Easy 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 length of the diameter of the tree . The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root . The length of a path between two nodes is represented by the number of edges between them. Example 1: Input: root = [1,2,3,4,5] Output: 3 Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3]. Example 2: Input: root = [1,2] Output: 1 Constraints: The number of no

Detailed Explanation

The problem asks us to find the diameter of a binary tree. The diameter is defined as the longest path between any two nodes in the tree. This path may or may not pass through the root node. The length of a path is the number of edges between the nodes. The input is the root node of the binary tree, and the output is the length of the diameter.

Solution Approach

The solution utilizes a recursive Depth-First Search (DFS) approach. The core idea is to calculate the height of the left and right subtrees for each node. During this height calculation, we also update the diameter of the tree. The diameter at a node is the sum of the left subtree's height and the right subtree's height. We maintain a global variable (or a member variable in a class) to keep track of the maximum diameter encountered so far. The DFS function returns the height of the subtree rooted at the current node.

Step-by-Step Algorithm

Step 1: Initialize a variable (e.g., `diameter` or `maxDiameter`) to store the maximum diameter found so far. Initialize it to 0.
Step 2: Define a recursive helper function (e.g., `depth` or `maxDepth`) that takes a node as input.
Step 3: Base case: If the node is NULL (or None), return 0 (height of an empty tree is 0).
Step 4: Recursively call the helper function on the left child to get the height of the left subtree (leftDepth).
Step 5: Recursively call the helper function on the right child to get the height of the right subtree (rightDepth).
Step 6: Update the `diameter` variable with the maximum of its current value and the sum of `leftDepth` and `rightDepth`. This step calculates the diameter that passes through the current node.
Step 7: Return the height of the subtree rooted at the current node, which is 1 + max(leftDepth, rightDepth). This value is used by the parent node's calculation.
Step 8: Call the helper function on the root node to start the DFS traversal.
Step 9: Return the value of the `diameter` variable, which now holds the maximum diameter found in the tree.

Key Insights

Insight 1: The diameter of a tree at a given node can be computed by summing the depths of its left and right subtrees.
Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the depth of each subtree. While calculating the depth, we can simultaneously update the diameter.
Insight 3: The diameter might not pass through the root, so we need to keep track of the maximum diameter found so far during the DFS traversal.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: Aurora, LinkedIn, Verkada, Visa, Wix.

Diameter of Binary Tree | LeetCode 543

Easy

Tree Depth-First Search Binary Tree

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

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;
 *     }
 * }
 */
class Solution {
    int maxDiameter = 0;

    public int diameterOfBinaryTree(TreeNode root) {
        maxDepth(root);
        return maxDiameter;
    }

    private int maxDepth(TreeNode node) {
        if (node == null) {
            return 0;
        }

        int leftDepth = maxDepth(node.left);
        int rightDepth = maxDepth(node.right);

        maxDiameter = Math.max(maxDiameter, leftDepth + rightDepth);

        return Math.max(leftDepth, rightDepth) + 1;
    }
}

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Problem Info

Description

Problem 543: Diameter of Binary Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

AuroraLinkedInVerkadaVisaWix

Stats

Difficulty:Easy

Languages:4

Companies:5

Problem #:543

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: Initialize a variable (e.g., `diameter` or `maxDiameter`) to store the maximum diameter found so far. Initialize it to 0.
2
Step 2: Define a recursive helper function (e.g., `depth` or `maxDepth`) that takes a node as input.
3
Step 3: Base case: If the node is NULL (or None), return 0 (height of an empty tree is 0).
4
Step 4: Recursively call the helper function on the left child to get the height of the left subtree (leftDepth).
5
Step 5: Recursively call the helper function on the right child to get the height of the right subtree (rightDepth).
6
Step 6: Update the `diameter` variable with the maximum of its current value and the sum of `leftDepth` and `rightDepth`. This step calculates the diameter that passes through the current node.
7
Step 7: Return the height of the subtree rooted at the current node, which is 1 + max(leftDepth, rightDepth). This value is used by the parent node's calculation.
8
Step 8: Call the helper function on the root node to start the DFS traversal.
9
Step 9: Return the value of the `diameter` variable, which now holds the maximum diameter found in the tree.

💡

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(h)

📝

Why?

The time complexity is O(n) because we visit each node once during the DFS traversal. The space complexity is determined by the maximum depth of the recursion call stack, which is proportional to the height of the tree.

📖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 DFS with Stack

The DFS can be implemented iteratively using a stack to avoid recursion. This can be beneficial in languages with limitations on recursion depth.

Time: O(n)Space: O(n) in the worst case (skewed tree), O(log n) in average case (balanced tree)

Trade-offs: Eliminates the risk of stack overflow compared to recursive approach but potentially more complex to implement.

✨

Key Insights

1Insight 1: The diameter of a tree at a given node can be computed by summing the depths of its left and right subtrees.
2Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the depth of each subtree. While calculating the depth, we can simultaneously update the diameter.
3Insight 3: The diameter might not pass through the root, so we need to keep track of the maximum diameter found so far during the DFS traversal.

📚

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 code handles this correctly in the base case of the recursive function, returning 0 and not affecting the diameter.
Edge case 2: Tree with only one node. The leftDepth and rightDepth will both be 0, so diameter will stay 0, which is the correct answer.
Edge case 3: Skewed tree (all nodes on one side). The height will be n, and the space complexity will be O(n).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to update the diameter within the recursive function. If the diameter is not updated at each node based on the sum of left and right depths, the correct maximum diameter may not be found.

✗Common mistake 2: Incorrectly calculating the height of the subtree. The height should be 1 + max(leftDepth, rightDepth), not just max(leftDepth, rightDepth).

✗Common mistake 3: Assuming the diameter always passes through the root node. The diameter can be between any two nodes in the tree.

✅

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 longest path between two devices in a network can be modeled as finding the diameter of a tree.

🏦Real-world scenario 2: Social networks. Determining the degree of separation between two individuals in a social network can be related to finding paths in a graph (which can sometimes be approximated as a tree structure).

🎮Real-world scenario 3: Phylogenetic tree analysis. Estimating the evolutionary distance between two species can be related to finding paths in a phylogenetic tree.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: June 4, 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

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

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Diameter of Binary Tree"?

For this easy-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?