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7. Symmetric Tree

easyAsked at Mercury

Determine if a binary tree is a mirror of itself around its center.

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

Problem

Given the root of a binary tree, check whether it is a mirror of itself. A tree is symmetric if its left subtree is a structural mirror of its right subtree with matching values.

Constraints

  • Node 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. Iterative pair queue

Enqueue mirrored pairs and compare values per pop.

Time
O(n)
Space
O(n)
const q=[[root.left,root.right]];
while(q.length){const [a,b]=q.shift();
  if(!a&&!b) continue; if(!a||!b||a.val!==b.val) return false;
  q.push([a.left,b.right],[a.right,b.left]);}
return true;

Tradeoff:

2. Mirror recursion

Recurse comparing left-of-left vs right-of-right and left-of-right vs right-of-left. Linear in nodes.

Time
O(n)
Space
O(h)
function isSymmetric(root) {
  const mirror = (a, b) => {
    if (!a && !b) return true;
    if (!a || !b || a.val !== b.val) return false;
    return mirror(a.left, b.right) && mirror(a.right, b.left);
  };
  return !root || mirror(root.left, root.right);
}

Tradeoff:

Mercury-specific tips

Mercury frames symmetry checks as verifying that ACH credit and debit halves of an internal book-transfer mirror each other — drift one side and reconciliation breaks.

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Output

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