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12. Symmetric Tree

easyAsked at Reddit

Determine if a binary tree is a mirror image of itself. Reddit asks this to gauge whether candidates can write paired-pointer recursion — the same skill needed when comparing two views of a comment tree (live vs. cached).

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

Source citations

Public interview reports confirming this problem appears in Reddit loops.

  • Glassdoor (2026-Q1)— Reddit comments-platform phone screen.

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 and check palindrome

Inorder-traverse, check if the resulting array is a palindrome.

Time
O(n)
Space
O(n)
function isSymmetric(root) {
  const vals = [];
  function inorder(n) {
    if (!n) { vals.push(null); return; }
    inorder(n.left); vals.push(n.val); inorder(n.right);
  }
  inorder(root);
  for (let i = 0, j = vals.length - 1; i < j; i++, j--) {
    if (vals[i] !== vals[j]) return false;
  }
  return true;
}

Tradeoff: Works but allocates a full traversal array. Subtle: null markers needed or {1,2,2,null,3,null,3} appears symmetric.

2. Paired recursion (optimal)

Compare left.left to right.right and left.right to right.left recursively.

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

Tradeoff: Clean, recursive, short-circuiting. The structural check (one null + one non-null) is crucial.

Reddit-specific tips

Reddit grades on the paired-recursion insight: you call mirror(left.left, right.right) and mirror(left.right, right.left). Many candidates try to compare left and right as a unit and confuse themselves.

Common mistakes

  • Forgetting to check structural equality (null vs non-null) — leads to comparing values on null nodes.
  • Calling mirror(left, right) on the same children: mirror(L.left, R.left) instead of mirror(L.left, R.right).
  • Iterative version: forgetting to push pairs of pointers (left-from-left, right-from-right).

Follow-up questions

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

  • Iterative version using a queue of pairs.
  • Same tree (LC 100) — the simpler cousin.
  • Invert binary tree (LC 226) — related transformation.

Solve it now

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Output

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FAQ

Why mirror(left.left, right.right)?

Symmetry means the leftmost of the left subtree mirrors the rightmost of the right subtree. The recursion encodes that geometric flip.

Can this be iterative?

Yes — use a queue, push pairs (left, right). Pop two at a time and check, then push (L.left, R.right) and (L.right, R.left).

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