What is the time complexity of Flip Equivalent Binary Trees?

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

What data structures are used to solve Flip Equivalent Binary Trees?

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

What are the key insights for solving Flip Equivalent Binary Trees?

Insight 1: Recursion is a natural fit for traversing tree structures and checking for equivalence at each level. Insight 2: We need to consider two possibilities at each node: the subtrees are either in the same order, or they need to be flipped to become equivalent.

Which companies ask LeetCode 951. Flip Equivalent Binary Trees in interviews?

This problem is frequently asked at Anduril.

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Flip Equivalent Binary Trees - LeetCode 951 Solution

Flip Equivalent Binary Trees - Complete Solution Guide

Flip Equivalent Binary Trees is LeetCode problem 951, 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

For a binary tree T , we can define a flip operation as follows: choose any node, and swap the left and right child subtrees. A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations. Given the roots of two binary trees root1 and root2 , return true if the two trees are flip equivalent or false otherwise. Example 1: Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7] Output:

Detailed Explanation

The problem asks us to determine if two binary trees, `root1` and `root2`, are 'flip equivalent'. Two trees are flip equivalent if we can make `root1` identical to `root2` by performing zero or more 'flip operations'. A flip operation involves selecting any node in a tree and swapping its left and right child subtrees. The input is the root nodes of the two binary trees, and the output is a boolean indicating whether they are flip equivalent.

Solution Approach

The solution utilizes a recursive approach to traverse both binary trees simultaneously. At each step, it checks for base cases. If the current nodes are both null, it means the trees are equivalent up to that point. If only one is null or their values differ, they aren't equivalent. Otherwise, it recursively checks if the left subtree of the first tree is flip equivalent to either the left or the right subtree of the second tree, and similarly for the right subtrees. The two recursive calls are combined using the logical OR operator, indicating that either the subtrees are in the original order, or one set of subtrees needs to be flipped.

Step-by-Step Algorithm

Step 1: Handle the base cases: If both `root1` and `root2` are null, return `true`. If one is null and the other isn't, or if their values differ, return `false`.
Step 2: Recursively check if `root1.left` is flip equivalent to `root2.left` AND `root1.right` is flip equivalent to `root2.right` (no flip needed).
Step 3: Recursively check if `root1.left` is flip equivalent to `root2.right` AND `root1.right` is flip equivalent to `root2.left` (flip needed).
Step 4: Return `true` if either the result from Step 2 or Step 3 is `true`. Otherwise, return `false`.

Key Insights

Insight 1: Recursion is a natural fit for traversing tree structures and checking for equivalence at each level.
Insight 2: We need to consider two possibilities at each node: the subtrees are either in the same order, or they need to be flipped to become equivalent.
Insight 3: Base cases are crucial: both trees being null are equivalent, and one tree being null while the other isn't, or differing root values, mean they're not equivalent.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(h)

Topics

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

Companies

Asked at: Anduril.

Flip Equivalent Binary Trees | LeetCode 951

Medium

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 {
    public boolean flipEquiv(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return true;
        }
        
        if (root1 == null || root2 == null || root1.val != root2.val) {
            return false;
        }
            
        // Check if subtrees are equivalent without flipping
        boolean no_flip_equiv = (flipEquiv(root1.left, root2.left) && 
                         flipEquiv(root1.right, root2.right));
        
        // Check if subtrees are equivalent with flipping
        boolean flip_equiv = (flipEquiv(root1.left, root2.right) && 
                      flipEquiv(root1.right, root2.left));
        
        return no_flip_equiv || flip_equiv;
    }
}

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Problem Info

Description

Problem 951: Flip Equivalent Binary Trees - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Binary Tree

Companies

Anduril

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:951

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

4 steps

1
Step 1: Handle the base cases: If both `root1` and `root2` are null, return `true`. If one is null and the other isn't, or if their values differ, return `false`.
2
Step 2: Recursively check if `root1.left` is flip equivalent to `root2.left` AND `root1.right` is flip equivalent to `root2.right` (no flip needed).
3
Step 3: Recursively check if `root1.left` is flip equivalent to `root2.right` AND `root1.right` is flip equivalent to `root2.left` (flip needed).
4
Step 4: Return `true` if either the result from Step 2 or Step 3 is `true`. Otherwise, return `false`.

💡

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, in the worst-case scenario, we have to visit each node of both trees once. The space complexity is O(h) due to the depth of the recursion stack. Each recursive call adds a new frame to the stack, and the maximum depth of the stack is equal 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 Approach with Stacks/Queues

While the recursive approach is elegant, an iterative approach using stacks or queues can also be used to traverse the trees. This approach avoids the overhead of recursion and can be useful if stack overflow is a concern.

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

Trade-offs: Iterative approach uses O(n) space in the worst case (when the stack/queue contains all nodes). The recursive approach uses O(h) stack space, where 'h' is the height of the tree. If the tree is balanced, the recursive approach is more space-efficient. However, the iterative approach avoids potential stack overflow issues. The code for the iterative approach is more complex than the recursive approach.

✨

Key Insights

1Insight 1: Recursion is a natural fit for traversing tree structures and checking for equivalence at each level.
2Insight 2: We need to consider two possibilities at each node: the subtrees are either in the same order, or they need to be flipped to become equivalent.
3Insight 3: Base cases are crucial: both trees being null are equivalent, and one tree being null while the other isn't, or differing root values, mean they're not equivalent.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty trees: If both trees are empty (`root1` and `root2` are null), the solution correctly returns `true` because empty trees are considered equivalent.
Edge case 2: One tree is empty, the other is not: If one tree is empty and the other is not, the solution returns `false`, indicating they are not flip equivalent.
Edge case 3: Different root values: If the root values of the two trees are different, the solution returns `false`, as the roots must be equal for the trees to be flip equivalent.
Edge case 4: Large trees where flips are necessary: The solution effectively handles scenarios requiring multiple flip operations by recursively checking both flipped and non-flipped configurations for the subtrees.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the base cases: Failing to handle the base cases (both trees null, one tree null, different root values) will lead to incorrect results or runtime errors.

✗Common mistake 2: Only checking for equivalence without flipping: Not considering the possibility that subtrees might need to be flipped can lead to the algorithm incorrectly reporting that trees are not equivalent when they actually are.

✗Common mistake 3: Not understanding the recursive calls: Confusing the recursive calls, particularly the 'no flip' and 'flip' equivalency checks, can result in an incorrect implementation.

✅

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: Data comparison and validation: Determining if two representations of hierarchical data (e.g., configuration files, XML documents) are equivalent, allowing for variations in ordering.

🏦Real-world scenario 2: Software testing: Verifying the correctness of tree transformations in compilers or other software tools, where the order of operations might be flexible but the final result must be consistent.

🎮Real-world scenario 3: Phylogenetic tree comparison: In biology, comparing phylogenetic trees, which represent the evolutionary relationships between species. The order of branches might be flipped without changing the underlying relationships.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: July 22, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

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

@CodeNinjaJun 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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Flip Equivalent Binary Trees"?

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?