Start by clarifying that symmetry means the left subtree is a mirror reflection of the right subtree. Then present both recursive and iterative solutions: the recursive one compares left and right subtrees using a helper function, while the iterative one uses a queue to compare nodes level by level in pairs.
Pro tip: Mention that the iterative solution can use a queue or stack, and that the recursive solution may hit stack overflow for very deep trees, so the iterative approach is often preferred in production. Also, note that an empty tree is symmetric.
Define symmetry: a tree is symmetric if the left subtree is a mirror image of the right subtree. Confirm edge cases: empty tree is symmetric, single node is symmetric.
Write a helper function isMirror(left, right) that returns true if both are null, or if both are non-null and their values are equal and isMirror(left.left, right.right) and isMirror(left.right, right.left) are true. Call isMirror(root, root) or isMirror(root.left, root.right).
Use a queue (or stack) to store pairs of nodes to compare. Initialize with (root, root) or (root.left, root.right). While the queue is not empty, dequeue two nodes, check if both null (continue), if one null or values differ (return false), then enqueue (left.left, right.right) and (left.right, right.left).
State that both solutions have O(n) time complexity and O(n) space complexity (recursive call stack or queue size). Mention that the iterative solution avoids stack overflow for deep trees.
Walk through a simple symmetric tree (e.g., [1,2,2,3,4,4,3]) and an asymmetric tree (e.g., [1,2,2,null,3,null,3]) to demonstrate correctness.
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