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11. Symmetric Tree

easyAsked at Asana

Determine if a binary tree is a mirror of itself. Asana asks this to test whether you can pair two pointers across a tree — the same pattern used in their snapshot-diff symmetry checks.

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

Source citations

Public interview reports confirming this problem appears in Asana loops.

  • Glassdoor (2026-Q1)— Asana tree warmup.

Problem

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center). Follow up: Could you solve it both recursively and iteratively?

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 serialize + palindrome check

Inorder traverse, build value list, check palindrome.

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

Tradeoff: Works but O(n) extra space and easy to get wrong with nulls.

2. Paired recursion

Recurse with (left, right). Mirror means left.left mirrors right.right and left.right mirrors right.left.

Time
O(n)
Space
O(h)
function isSymmetric(root) {
  const 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 !root || mirror(root.left, root.right);
}

Tradeoff: O(h) stack, no extra storage. The crossed-recursion (left.left vs right.right, left.right vs right.left) is the trick.

Asana-specific tips

Asana cares about the 'crossed' recursion pattern — they grade on whether you immediately see (left.left, right.right) and (left.right, right.left) as the mirror invariant. Drawing two trees side-by-side on the whiteboard before coding earns interviewer trust.

Common mistakes

  • Calling mirror(a.left, b.left) — that's same-tree comparison, not mirror.
  • Forgetting the root === null case.
  • Returning early on (a.val === b.val) without recursing into children.

Follow-up questions

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

  • Iterative version using a queue of pairs.
  • Count the size of the largest symmetric subtree.
  • Mirror a binary tree (LC 226).

Solve it now

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Output

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FAQ

Why is left.left compared to right.right?

Mirroring means flipping. The leftmost node on the left subtree corresponds to the rightmost node on the right subtree — that's left.left mirroring right.right.

Does this work iteratively?

Yes — use a queue (or stack) of pairs. Push (root.left, root.right) and BFS, pushing children in the mirrored order each time.

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