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11. Symmetric Tree

easyAsked at Slack

Check whether a binary tree is a mirror of itself.

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

Problem

Given the root of a binary tree, check whether it is a mirror of itself (symmetric around its center).

Constraints

  • Nodes count in [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 + check palindrome

Inorder traverse with null markers, check if it's a palindrome.

Time
O(n)
Space
O(n)
const arr=[];
function dfs(n){ if(!n){arr.push(null);return;} dfs(n.left);arr.push(n.val);dfs(n.right);}
dfs(root);
return arr.join(',')===arr.reverse().join(',');

Tradeoff:

2. Mirror recursion

Recurse on (left.left, right.right) and (left.right, right.left). Treat null mirrors as equal.

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 !root || mirror(root.left, root.right);
}

Tradeoff:

Slack-specific tips

Slack interviewers like the mirror-recursion pattern because it generalizes to UI-tree comparison (RTL vs LTR layouts) — call that out.

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Output

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