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22. Symmetric Tree

easyAsked at GitHub

Check whether a binary tree is a mirror of itself using recursive or iterative paired traversal, a BFS/DFS tree skill GitHub tests in screening rounds.

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

  • The number of nodes is in 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. Serialize and string compare

Serialize left and right subtrees then compare — verbose and allocates O(n) strings unnecessarily.

Time
O(n)
Space
O(n)
// serialize(left) === serialize(mirror of right)
// Works but extra allocation and code complexity

Tradeoff:

2. Recursive mirror check

Define a helper isMirror(left, right) that checks outer pairs (left.left vs right.right) and inner pairs (left.right vs right.left) recursively.

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

Tradeoff:

GitHub-specific tips

GitHub screens with this problem to verify clean recursive thinking; also be ready with the iterative BFS queue variant pairing nodes level-by-level, since interviewers often ask for both.

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Output

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