Symmetric Tree

Problem

Given the root of a binary tree, decide whether it is a mirror image of itself (symmetric around its centre).

Constraints

• 1 ≤ nodes ≤ 1000

Approach — Tree Traversal

This is a Tree Traversal problem. The idea: traverse the tree — DFS recursively, or BFS level by level — and combine the results from each subtree. Work through the reference code below line by line, then re-derive it yourself in the editor — that's how the pattern sticks.

Complexity: O(n) time.

Solution code

Python

class Solution:
    def isSymmetric(self, root):
        def mirror(a, b):
            if not a and not b:
                return True
            if not a or not b or a.val != b.val:
                return False
            return mirror(a.left, b.right) and mirror(a.right, b.left)
        return mirror(root.left, root.right) if root else True

Java

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return root == null || mirror(root.left, root.right);
    }
    private boolean mirror(TreeNode a, TreeNode b) {
        if (a == null && b == null) return true;
        if (a == null || b == null) return false;
        return a.val == b.val && mirror(a.left, b.right) && mirror(a.right, b.left);
    }
}

Practice it

Reading a solution isn't the same as being able to write it under pressure. Open this problem in the in-browser editor, hide the solution, and solve it from scratch — your code runs against real test cases instantly.