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11. Symmetric Tree

easyAsked at Vercel

Given a binary tree, check if it is a mirror of itself (symmetric around its center). Vercel asks this because the recursive mirror-comparison shape shows up in their A/B route-mirror experiments.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Source citations

Public interview reports confirming this problem appears in Vercel loops.

  • Glassdoor (2025-Q4)— Vercel screen recursion warm-up.
  • Blind (2026-Q1)— Mentioned in Vercel onsite recap as a quick tree question.

Problem

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

Constraints

  • The number of nodes in the tree is in the range [1, 1000].
  • -100 <= Node.val <= 100

Examples

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

Approaches

1. Inorder traversal + palindrome check

Inorder traverse, then check the resulting array is a palindrome.

Time
O(n)
Space
O(n)
function isSymmetric(root) {
  // Caveat: nulls must be encoded to distinguish [1,2,null] from [1,null,2].
  const arr = [];
  function dfs(n) { if (!n) { arr.push(null); return; } dfs(n.left); arr.push(n.val); dfs(n.right); }
  dfs(root);
  // palindrome check on arr...
}

Tradeoff: Tricky to get null-encoding right; the recursive mirror is cleaner.

2. Recursive mirror comparison (optimal)

Helper compares (left, right). Both null → true. One null → false. Values equal AND left.left mirrors right.right AND left.right mirrors right.left.

Time
O(n)
Space
O(h)
function isSymmetric(root) {
  function mirror(a, b) {
    if (!a && !b) return true;
    if (!a || !b) return false;
    return a.val === b.val
        && mirror(a.left, b.right)
        && mirror(a.right, b.left);
  }
  return mirror(root.left, root.right);
}

Tradeoff: The mirror pattern (left.left vs right.right, left.right vs right.left) is the whole trick. Articulate it explicitly.

Vercel-specific tips

Vercel wants you to articulate the mirror pattern out loud BEFORE coding: outer-vs-outer, inner-vs-inner. Bonus signal: offering the iterative BFS version with a queue of pairs as an alternative.

Common mistakes

  • Comparing left.left to right.left — that's structural identity, not mirror.
  • Forgetting the null pair check — crashes on mid-tree nulls.
  • Treating an empty tree as not symmetric — empty IS symmetric; check the constraints.

Follow-up questions

An interviewer at Vercel may pivot to one of these next:

  • Iterative version using a queue of (left, right) pairs.
  • Invert a binary tree (LC 226) — related shape.
  • Symmetric N-ary tree — generalize the mirror to k children.

Solve it now

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Output

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FAQ

Why outer-vs-outer and inner-vs-inner?

Mirroring means the leftmost descendant on the left subtree should match the rightmost descendant on the right subtree. That's left.left vs right.right (outer) and left.right vs right.left (inner).

Can I solve this iteratively?

Yes — enqueue (root.left, root.right) and process pairs. Same logic in BFS form. Useful if recursion depth is a worry.

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