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6. Symmetric Tree

easyAsked at Flipkart

Decide whether a binary tree is a mirror of itself — Flipkart uses it as a quick recursion warm-up before moving to inventory-tree questions.

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). Return true if symmetric.

Constraints

  • 1 <= nodes <= 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. Brute force level capture

BFS each level into an array and check palindrome with nulls preserved.

Time
O(n)
Space
O(n)
// level-order with nulls, then for each level check palindrome
// extra memory, easy bug to drop nulls

Tradeoff:

2. Mirror DFS

Recurse on (left.left, right.right) and (left.right, right.left) in lockstep. Constant extra work per node.

Time
O(n)
Space
O(h)
function isSymmetric(root) {
  const 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:

Flipkart-specific tips

Flipkart panels reward candidates who call out the null-vs-value asymmetry trap — it parallels their nullable attribute checks in category metadata.

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Output

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