What is the time complexity of Merge Two Binary Trees?

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

What data structures are used to solve Merge Two Binary Trees?

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

What are the key insights for solving Merge Two Binary Trees?

Insight 1: Recursion is a natural fit because the merging process is applied to each node and its subtrees. Insight 2: We can modify the `root1` tree directly to create the merged tree, saving space. If either root is null, the other root (or null if both are null) should be returned directly.

Which companies ask LeetCode 617. Merge Two Binary Trees in interviews?

This problem is frequently asked at MongoDB.

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Merge Two Binary Trees - LeetCode 617 Solution

Merge Two Binary Trees - Complete Solution Guide

Merge Two Binary Trees is LeetCode problem 617, 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

You are given two binary trees root1 and root2 . Imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not. You need to merge the two trees into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of the new tree. Return the merged tree . Note: The merging process must start from the root nodes of both t

Detailed Explanation

The problem asks us to merge two binary trees, `root1` and `root2`, into a single new binary tree. If two nodes in the same position in both trees exist (overlap), their values should be summed to create the value of the corresponding node in the merged tree. If a node only exists in one of the trees, that node becomes part of the merged tree. The merging process starts from the root nodes of both trees.

Solution Approach

The solution uses a recursive approach. It first checks if either `root1` or `root2` is null. If `root1` is null, `root2` is returned (or null if `root2` is also null). If `root2` is null, `root1` is returned. If both are not null, the value of `root2` is added to `root1`. Then, the left and right subtrees are merged recursively and assigned to the left and right children of `root1` respectively. Finally, `root1` is returned, which is now the merged tree.

Step-by-Step Algorithm

Step 1: Check if either `root1` or `root2` is null. If `root1` is null, return `root2`. If `root2` is null, return `root1`.
Step 2: If both `root1` and `root2` are not null, add the value of `root2`'s node to `root1`'s node.
Step 3: Recursively call `mergeTrees` with the left subtrees of `root1` and `root2` and assign the result to the left child of `root1`.
Step 4: Recursively call `mergeTrees` with the right subtrees of `root1` and `root2` and assign the result to the right child of `root1`.
Step 5: Return `root1`. `root1` now represents the merged tree.

Key Insights

Insight 1: Recursion is a natural fit because the merging process is applied to each node and its subtrees.
Insight 2: We can modify the `root1` tree directly to create the merged tree, saving space. If either root is null, the other root (or null if both are null) should be returned directly.
Insight 3: The base cases for the recursion are when either `root1` or `root2` is null. This handles cases where one tree is larger than the other or when we reach the end of a branch.

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: MongoDB.

Merge Two Binary Trees | LeetCode 617

Easy

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;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if(root1 == null) return root2;
        if(root2 == null) return root1;
        root1.val += root2.val;
        root1.left = mergeTrees(root1.left, root2.left);
        root1.right = mergeTrees(root1.right, root2.right);
        return root1;
    }
}

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Problem Info

Description

Problem 617: Merge Two Binary Trees - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

MongoDB

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:617

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

5 steps

1
Step 1: Check if either `root1` or `root2` is null. If `root1` is null, return `root2`. If `root2` is null, return `root1`.
2
Step 2: If both `root1` and `root2` are not null, add the value of `root2`'s node to `root1`'s node.
3
Step 3: Recursively call `mergeTrees` with the left subtrees of `root1` and `root2` and assign the result to the left child of `root1`.
4
Step 4: Recursively call `mergeTrees` with the right subtrees of `root1` and `root2` and assign the result to the right child of `root1`.
5
Step 5: Return `root1`. `root1` now represents the merged 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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because each node in both trees is visited at most once during the merge operation. The space complexity is O(n) due to the recursive call stack. The depth of the recursion depends on the height of the trees. In the worst-case scenario (skewed trees), the height can be n, leading to O(n) space complexity. If the trees were balanced the space complexity would reduce to O(log n).

📖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 Approach (Breadth-First Search)

Instead of recursion, an iterative solution using a queue can be implemented. Use a queue to store pairs of nodes from `root1` and `root2` to be merged. The merging logic is the same, but implemented iteratively.

Time: O(n), where n is the minimum number of nodes in the two trees.Space: O(n), where n is the maximum number of nodes at any level in the trees.

Trade-offs: Iterative approach might be preferred if stack overflow is a concern due to extremely deep trees. However, the recursive approach can be more concise and easier to read in this case.

✨

Key Insights

1Insight 1: Recursion is a natural fit because the merging process is applied to each node and its subtrees.
2Insight 2: We can modify the `root1` tree directly to create the merged tree, saving space. If either root is null, the other root (or null if both are null) should be returned directly.
3Insight 3: The base cases for the recursion are when either `root1` or `root2` is null. This handles cases where one tree is larger than the other or when we reach the end of a branch.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Both trees are empty (null). The code correctly returns null, as both `root1` and `root2` will be null in the initial check.
Edge case 2: One tree is much larger than the other. The recursion will continue down the larger tree, appending the remaining part of the larger tree to the merged tree.
Edge case 3: One tree has a node while the other doesn't. The non-null node is included in the result.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the base cases (null roots) correctly, leading to null pointer exceptions or incorrect merging.

✗Common mistake 2: Creating new nodes instead of modifying the nodes of the first tree (`root1`), leading to unnecessary memory allocation. The problem description specifies modifying root1 in place.

✗Common mistake 3: Forgetting to recursively merge the left and right subtrees, resulting in only merging the root nodes.

✅

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: Combining different versions of a configuration file represented as a tree structure. For example, merging user-specific settings with default settings.

🏦Real-world scenario 2: Image blending where each pixel is represented by a node in a tree structure (e.g., quadtree). Merging two images would involve combining pixel values in overlapping regions.

💡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

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ 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: August 17, 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

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

@TechLeadJun 30

❓

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 "Merge Two Binary Trees"?

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?