DSA TrainerSymmetric Tree
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
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Problem
Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).
Input
The root of a binary tree.
Output
`true` if the tree is symmetric, `false` otherwise.
Examples
Input: root = [1,2,2,3,4,4,3]
Output: true
The left subtree and right subtree are mirror images: left.val=2, right.val=2; left-left.val=3 mirrors right-right.val=3; left-right.val=4 mirrors right-left.val=4.
Input: root = [1,2,2,null,3,null,3]
Output: false
The right children of both level-2 nodes are 3, but a mirror requires them on opposite sides.
The brute-force approach
Collect the level-order traversal and check each level to see if it's a palindrome.
# BFS level by level, check each level is a palindrome
# Must include nulls to detect structural asymmetryO(n) time and O(n) space for the BFS queue — same complexity as the recursive approach, but more code and null tracking is tricky to get right.
Spotting the pattern
This is a Tree Traversal problem. The key question to ask yourself:
What does it mean for two subtrees to be mirrors of each other?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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