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12. Symmetric Tree

easyAsked at Workday

Given the root of a binary tree, check whether it is a mirror of itself. Workday uses this to extend the parallel-recursion pattern from Same Tree — useful for symmetrical permission policies in dual-approval workflows.

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

Source citations

Public interview reports confirming this problem appears in Workday loops.

  • Glassdoor (2025)— Workday SDE2 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 serialize + reverse check

Inorder-serialize with null markers, check palindrome.

Time
O(n)
Space
O(n)
// serialize with explicit nulls, then compare with reversed serialization
// fragile — equal values on same depth on opposite sides may collide

Tradeoff: Brittle and slow. Don't.

2. Mirror-pair recursion

Compare left.left to right.right AND left.right to right.left.

Time
O(n)
Space
O(h) recursion
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, root);
}

Tradeoff: Same as Same Tree but with the swap — left.left vs right.right is the only structural difference.

Workday-specific tips

Workday wants you to articulate that this is Same Tree with a swap. The cross-recursion (left.left <-> right.right) is the entire insight. Bonus signal: mention iterative version with a queue of pairs.

Common mistakes

  • Comparing left to right at the top — that's same tree, not symmetric.
  • Forgetting one cross (e.g., only doing left.left <-> right.right but not left.right <-> right.left).
  • Starting with mirror(root.left, root.right) instead of mirror(root, root) — works but loses the elegant pattern.

Follow-up questions

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

  • Iterative with a queue.
  • Same Tree (LC 100) — same pattern, no swap.
  • Invert Binary Tree (LC 226).

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Output

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FAQ

Why mirror(root, root) instead of mirror(root.left, root.right)?

Both work. mirror(root, root) is more uniform — the function compares two trees, and the special case is just that they start identical.

How would I do this iteratively?

Use a queue. Push (root, root). Pop pairs; on each, push (a.left, b.right) and (a.right, b.left). Fail on mismatched structure or values.

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