What is the time complexity of Symmetric Tree?

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

What data structures are used to solve Symmetric Tree?

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 Symmetric Tree?

**Coordinated Symmetric Traversal:** Instead of a single standard tree traversal, this problem demands a coordinated traversal pattern. We're effectively running two 'virtual pointers' (`t1` and `t2`) simultaneously. When `t1` explores a `left` child, `t2` must explore a `right` child to maintain the mirror comparison, and vice versa. This `t1.left` vs `t2.right` and `t1.right` vs `t2.left` logic is the bedrock of symmetric checks. **Recursive Definition of Mirroring:** The problem lends itself perfectly to recursion because the definition of a symmetric tree is inherently recursive. A tree is symmetric if its root's children's subtrees are mirror images of each other. The `isMirror` function directly translates this by ensuring the values match at the current level and then recursively confirming the mirrored relationship between the respective children.

Which companies ask LeetCode 101. Symmetric Tree in interviews?

This problem is frequently asked at Adobe, Apple, Bloomberg, Comcast, Google and 5 more companies.

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Symmetric Tree - LeetCode 101 Solution

Q: How to solve Symmetric Tree?

The provided solution employs a clever recursive helper function, `isMirror(t1, t2)`, to check if two subtrees (`t1` and `t2`) are mirror images of each other. The core idea is to initiate this check by calling `isMirror(root, root)`. This sets up the problem as comparing the tree against itself for mirroring, where the 'left half' will be traversed using `t1` and the 'right half' using `t2` (conceptually starting from `root.left` and `root.right` respectively in subsequent calls). The `isMirror` function then recursively validates three conditions: 1) if both `t1` and `t2` are null, they are symmetric; 2) if only one is null, they are not symmetric; 3) if both exist, their values must be equal, AND `t1.left` must mirror `t2.right`, AND `t1.right` must mirror `t2.left`. This last step is crucial as it enforces the cross-comparison required for symmetry.

Symmetric Tree - Complete Solution Guide

Symmetric Tree is LeetCode problem 101, 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, check whether it is a mirror of itself (i.e., symmetric around its center). Example 1: Input: root = [1,2,2,3,4,4,3] Output: true Example 2: Input: root = [1,2,2,null,3,null,3] Output: false Constraints: The number of nodes in the tree is in the range [1, 1000] . -100 <= Node.val <= 100 Follow up: Could you solve it both recursively and iteratively?

Detailed Explanation

This problem asks us to determine if a given binary tree is a mirror image of itself when reflected across its vertical center. It's not just about node values being identical at the same level; the structure and values must be perfectly symmetrical. Imagine drawing a line straight down the middle of the tree – the left side should look exactly like the right side, but flipped.

Solution Approach

The provided solution employs a clever recursive helper function, `isMirror(t1, t2)`, to check if two subtrees (`t1` and `t2`) are mirror images of each other. The core idea is to initiate this check by calling `isMirror(root, root)`. This sets up the problem as comparing the tree against itself for mirroring, where the 'left half' will be traversed using `t1` and the 'right half' using `t2` (conceptually starting from `root.left` and `root.right` respectively in subsequent calls). The `isMirror` function then recursively validates three conditions: 1) if both `t1` and `t2` are null, they are symmetric; 2) if only one is null, they are not symmetric; 3) if both exist, their values must be equal, AND `t1.left` must mirror `t2.right`, AND `t1.right` must mirror `t2.left`. This last step is crucial as it enforces the cross-comparison required for symmetry.

Step-by-Step Algorithm

Step 1: Check for empty tree: If the root is null, return `true` (an empty tree is symmetric).
Step 2: Call the recursive helper function `isMirror` with the root's left and right subtrees as arguments.
Step 3: In `isMirror` function: Check the base cases (both subtrees null, one subtree null, or values different).
Step 4: Recursively call `isMirror` on mirrored children (left's left to right's right, and left's right to right's left).
Step 5: Return `true` only if all recursive calls and value checks return `true`, indicating perfect mirroring.

Key Insights

**Coordinated Symmetric Traversal:** Instead of a single standard tree traversal, this problem demands a coordinated traversal pattern. We're effectively running two 'virtual pointers' (`t1` and `t2`) simultaneously. When `t1` explores a `left` child, `t2` must explore a `right` child to maintain the mirror comparison, and vice versa. This `t1.left` vs `t2.right` and `t1.right` vs `t2.left` logic is the bedrock of symmetric checks.
**Recursive Definition of Mirroring:** The problem lends itself perfectly to recursion because the definition of a symmetric tree is inherently recursive. A tree is symmetric if its root's children's subtrees are mirror images of each other. The `isMirror` function directly translates this by ensuring the values match at the current level and then recursively confirming the mirrored relationship between the respective children.
**Precise Null Node Handling:** The base cases `if not t1 and not t2: return True` and `if not t1 or not t2: return False` are more than just exit conditions; they are fundamental to correctly identifying symmetry. Two null branches are trivially symmetric, while a null branch paired with an existing node signifies an immediate break in symmetry. Correctly managing these null scenarios ensures that trees with missing nodes (like `[1,2,2,null,3,3,null]`) are evaluated accurately without crashing or producing false positives.

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: Adobe, Apple, Bloomberg, Comcast, Google, LinkedIn, Meta, Microsoft, Yahoo, Yandex.

Symmetric Tree | LeetCode 101

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 boolean isSymmetric(TreeNode root) {
        return isMirror(root, root);
    }

    private boolean isMirror(TreeNode t1, TreeNode t2) {
        if (t1 == null && t2 == null) return true;
        if (t1 == null || t2 == null) return false;
        return (t1.val == t2.val)
            && isMirror(t1.right, t2.left)
            && isMirror(t1.left, t2.right);
    }
}

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Problem Info

Description

Problem 101: Symmetric Tree - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Binary Tree

Companies

AdobeAppleBloombergComcastGoogleLinkedInMetaMicrosoft

Stats

Difficulty:Easy

Languages:4

Companies:10

Problem #:101

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 for empty tree: If the root is null, return `true` (an empty tree is symmetric).
2
Step 2: Call the recursive helper function `isMirror` with the root's left and right subtrees as arguments.
3
Step 3: In `isMirror` function: Check the base cases (both subtrees null, one subtree null, or values different).
4
Step 4: Recursively call `isMirror` on mirrored children (left's left to right's right, and left's right to right's left).
5
Step 5: Return `true` only if all recursive calls and value checks return `true`, indicating perfect mirroring.

💡

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 algorithm visits each node in the tree at most once. In the worst-case scenario (a completely skewed tree), the recursive depth is equal to the height of the tree, resulting in O(h) space complexity. For a balanced tree, h = log₂(n), making the space complexity O(log₂(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 (BFS)

Use a queue for Breadth-First Search to compare nodes level by level. This avoids recursion, potentially reducing the risk of stack overflow errors with very deep trees.

Time: O(n)Space: O(w), where w is the maximum width of the tree.

Trade-offs: Iterative approach offers slightly better space complexity in the average case for balanced trees, but it's less concise and potentially more verbose than the recursive version. The recursive approach is generally preferred for its elegance and readability in this particular problem.

✨

Key Insights

1**Coordinated Symmetric Traversal:** Instead of a single standard tree traversal, this problem demands a coordinated traversal pattern. We're effectively running two 'virtual pointers' (`t1` and `t2`) simultaneously. When `t1` explores a `left` child, `t2` must explore a `right` child to maintain the mirror comparison, and vice versa. This `t1.left` vs `t2.right` and `t1.right` vs `t2.left` logic is the bedrock of symmetric checks.
2**Recursive Definition of Mirroring:** The problem lends itself perfectly to recursion because the definition of a symmetric tree is inherently recursive. A tree is symmetric if its root's children's subtrees are mirror images of each other. The `isMirror` function directly translates this by ensuring the values match at the current level and then recursively confirming the mirrored relationship between the respective children.
3**Precise Null Node Handling:** The base cases `if not t1 and not t2: return True` and `if not t1 or not t2: return False` are more than just exit conditions; they are fundamental to correctly identifying symmetry. Two null branches are trivially symmetric, while a null branch paired with an existing node signifies an immediate break in symmetry. Correctly managing these null scenarios ensures that trees with missing nodes (like `[1,2,2,null,3,3,null]`) are evaluated accurately without crashing or producing false positives.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree. The solution correctly handles this by returning `true` in `isSymmetric` if the root is `null`. This implies symmetry as the tree is empty.
Edge case 2: Tree with only one node. This is correctly considered symmetric because there are no subtrees to compare.
Edge case 3: Trees with null children. The solution gracefully handles these through the base cases in `isMirror`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect handling of base cases (null subtrees). Failing to correctly manage situations where one or both subtrees are null leads to incorrect results.

✗Common mistake 2: Mixing up left and right child comparisons in the recursion. The mirroring must be precise: left subtree's left child compares to right subtree's right child, and vice versa. Any mix-up will result in a wrong answer.

✅

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: Checking for symmetry in data structures that represent mirrored or palindromic structures, such as certain types of network topologies or organizational charts.

🏦Real-world scenario 2: Validating the structural integrity of data within a tree where the symmetric structure is a requirement for correct function (e.g., certain types of cryptographic tree structures).

💡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: September 10, 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.

@AlgoMasterJun 24

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

@InterviewProJun 17

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

@CodeNinjaJun 10

❓

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 "Symmetric 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.

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