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11. Symmetric Tree

easyAsked at ServiceNow

Given the root of a binary tree, check whether it is a mirror of itself. ServiceNow asks this to confirm you can write a paired recursion (compare left.left with right.right, left.right with right.left) — useful in any mirrored workflow validation.

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

Source citations

Public interview reports confirming this problem appears in ServiceNow loops.

  • Glassdoor (2026-Q1)— ServiceNow loops use this to test paired-pointer recursion.
  • LeetCode Discuss (2025-12)— ServiceNow phone screen rotation includes this with explicit BFS follow-up.

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. Compare with reversed clone

Build a mirror clone, then compare equality (LC 100).

Time
O(n)
Space
O(n)
function isSymmetric(root) {
  const mirror = (n) => !n ? null : { val: n.val, left: mirror(n.right), right: mirror(n.left) };
  const same = (a,b) => (!a && !b) || (a && b && a.val === b.val && same(a.left,b.left) && same(a.right,b.right));
  return same(root, mirror(root));
}

Tradeoff: Allocates a whole second tree. Mention but show why ServiceNow prefers the paired-pointer recursion.

2. Paired recursion

Define isMirror(a, b). Base cases handle null. Recurse: a.left vs b.right AND a.right vs b.left.

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

Tradeoff: Linear time, no extra tree. The trick is mirroring the recursion: left.left pairs with right.right, left.right pairs with right.left.

ServiceNow-specific tips

ServiceNow grades for the paired recursion mirror pattern — they want isMirror(a.left, b.right) AND isMirror(a.right, b.left). Bonus signal: state the invariant 'two subtrees are mirrors iff their roots are equal and their left/right children mirror diagonally' before coding.

Common mistakes

  • Pairing left.left with right.left instead of right.right — symmetric error that flips the meaning.
  • Returning true when one is null but not the other.
  • Comparing whole subtrees as equal — that's same-tree, not mirror-tree.

Follow-up questions

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

  • Iterative BFS version using a queue of pairs.
  • Tree Diameter (LC 543).
  • Detecting symmetric subgraphs in an arbitrary graph.

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Output

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FAQ

Why call isMirror(root, root) and not isMirror(root.left, root.right)?

Calling on (root, root) lets the base case handle null root uniformly. Calling on the children would need a separate null check on root.

Can I do it iteratively?

Yes — use a queue. Push pairs (a, b). Each step pop a pair, compare, then push (a.left, b.right) and (a.right, b.left).

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