Is Symmetric Tree easy or hard?

Symmetric Tree is categorized as an Easy problem on most coding platforms with an acceptance rate around 60%. The challenge lies in recognizing the mirror comparison pattern rather than performing a standard traversal.

Symmetric Tree Python/Java solution

In Python or Java, the typical solution defines a helper function that recursively compares two nodes. The function checks value equality and recursively validates mirrored children. Both languages implement the same O(n) DFS logic with recursion.

How to solve Symmetric Tree in O(n)?

Traverse the tree while comparing nodes in mirrored positions. In the recursive solution, compare left.left with right.right and left.right with right.left at every step. Each node pair is processed once, giving O(n) time complexity with O(h) auxiliary space for recursion.

What is the best approach for Symmetric Tree?

The recursive mirror check using Depth-First Search is the most common solution. It compares two subtrees at a time and ensures the left subtree mirrors the right subtree. The algorithm runs in O(n) time because every node is visited once, and it uses O(h) space from the recursion stack where h is the tree height.

Is Symmetric Tree asked at Google/Amazon/Meta?

Symmetric Tree is a common entry-level tree question frequently asked in technical interviews at companies like Amazon, Meta, and Microsoft. It tests understanding of tree traversal, recursion, and mirror structure validation in binary trees.

What data structure is used in Symmetric Tree?

The problem uses a binary tree as the primary data structure. Solutions typically apply Depth-First Search with recursion or Breadth-First Search using a queue to compare mirrored nodes across the tree.

What is the time complexity of Symmetric Tree?

The optimal time complexity is O(n), where n is the number of nodes in the binary tree. Every node must be examined at least once to verify the mirror structure. Both recursive DFS and iterative BFS approaches achieve this complexity.

Description Solution Video 6:40

101. Symmetric Tree

Easy62.1% acceptance11 companies

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?

Asked at 11 companies

Meta LinkedIn VK Amazon Microsoft Bloomberg Yandex

Google Tiktok Apple Goldman Sachs

Topics

6:40

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1# Definition for a binary tree node.
2# class TreeNode:
3#     def __init__(self, val=0, left=None, right=None):
4#         self.val = val
5#         self.left = left
6#         self.right = right
7class Solution:
8    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
9        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2

Input

[1,2,2,3,4,4,3]

Input

[1,2,2,null,3,null,3]

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