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8. Symmetric Tree

easyAsked at Notion

Determine whether a binary tree is a mirror of itself. Notion likes this for the dual-pointer recursion that introduces tree-symmetry thinking.

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

Source citations

Public interview reports confirming this problem appears in Notion loops.

  • Glassdoor (2026-Q1)— Notion uses it to test mirror recursion.
  • Blind (2025-10)— Reported in onsite tree round.

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. BFS level + palindrome check

Collect each level (including nulls) into an array, check if it's a palindrome.

Time
O(n)
Space
O(w)
function isSymmetric(root) {
  let q = [root];
  while (q.length) {
    const vals = q.map(n => n ? n.val : null);
    let i=0, j=vals.length-1;
    while (i < j) { if (vals[i] !== vals[j]) return false; i++; j--; }
    const next = [];
    for (const n of q) if (n) { next.push(n.left); next.push(n.right); }
    q = next;
  }
  return true;
}

Tradeoff: Works but more bookkeeping than the dual recursion.

2. Dual recursion (optimal)

Compare left.left with right.right AND left.right with right.left. Both null = symmetric here; one null = asymmetric; values differ = asymmetric.

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: Clean and tiny. The cross-pairing (left-right with right-left) is the key insight.

Notion-specific tips

Notion grades whether you immediately see this as 'Same Tree, but with cross-pairing.' If you reuse the LC 100 pattern and just flip the recursive arguments, that's the cleanest sign. Don't write a separate iterative solution unless asked.

Common mistakes

  • Comparing left subtree to itself instead of cross-pairing with right.
  • Forgetting the both-null base case.
  • Trying to handle the root specially — start the recursion at (root.left, root.right).

Follow-up questions

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

  • Same Tree (LC 100).
  • Mirror the tree in-place (Invert Tree, LC 226).
  • What if it's an N-ary tree?

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Output

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FAQ

Why cross-pair the children?

Mirror symmetry means the leftmost child of the left subtree should equal the rightmost child of the right subtree. That's the cross pattern.

Iterative is asked — what's the trick?

Use a queue, but enqueue PAIRS of nodes (the two cross-paired nodes). Dequeue a pair, compare, then enqueue 4 sub-pairs.

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