Check if Two Binary Trees are Mirror Images - Algorithm, Visualization, Examples

Problem Statement❯

Given two binary trees, your task is to determine whether they are mirror images of each other. Two trees are said to be mirrors if the structure of one tree is the reverse of the other and their corresponding node values are the same. This check must be done recursively for all corresponding nodes.

Examples❯

Tree 1	Tree 2	Are Mirror?	Description
[1, 2, 3, 4, 5]	[1, 3, 2, null, null, 5, 4]	true	Tree 2 is the mirror image of Tree 1: left and right children are swapped at every node.
[10, 20, 30]	[10, 30, 20]	true	Left and right subtrees are perfectly mirrored.
[1, 2, 3]	[1, 2, 3]	false	Both trees have the same structure, but are not mirrors (no left-right reversal).
[]	[]	true	Two empty trees are trivially mirrors of each other.
[1]	[1]	true	Single-node trees with the same value are mirrors.
[1, 2]	[1, null, 2]	true	Tree 1 has a left child; Tree 2 has a right child. Structures mirror each other.

Solution❯

Understanding the Problem

We are given two binary trees and asked to determine whether they are mirror images of each other. This means that the structure and values of one tree must exactly reflect those of the other, but in a mirrored way. Specifically, for every node in the first tree, its left child must match the right child of the corresponding node in the second tree, and vice versa.

This problem is common in recursion-based tree problems and tests your ability to compare tree structures and values symmetrically.

Step-by-Step Solution with Example

Step 1: Understand the Mirror Concept

Two trees are mirrors if:

They are both empty (null)
Or, their root nodes have equal values, and their left and right subtrees are mirrors of each other in opposite directions

Step 2: Use a Recursive Approach

The simplest and cleanest way to solve this is using recursion. We define a function isMirror(tree1, tree2) that checks:

If both trees are null, return true
If only one is null, return false
If values don't match, return false
Recursively check isMirror(tree1.left, tree2.right) and isMirror(tree1.right, tree2.left)

Step 3: Apply the Approach to an Example

Let’s take two trees:


Tree A         Tree B
   1              1
  /             /  2   3          3   2

We compare:

Roots: 1 == 1 ✅
Left of A (2) with Right of B (2) ✅
Right of A (3) with Left of B (3) ✅

All checks pass recursively, so these trees are mirror images.

Step 4: Implement the Code


boolean isMirror(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) return true;
    if (t1 == null || t2 == null) return false;
    if (t1.val != t2.val) return false;
    return isMirror(t1.left, t2.right) && isMirror(t1.right, t2.left);
}

Edge Cases

Case 1: Both Trees Are Empty

Two null trees are mirrors by definition. Return true.

Case 2: One Tree is Empty

One null and one non-null tree can never be mirrors. Return false.

Case 3: Different Root Values

If the roots are not equal, trees can't be mirrors. Check t1.val != t2.val.

Case 4: Matching Roots, Unmatched Subtrees

Even if roots match, if the left and right children don't mirror each other in structure and value, return false.

Case 5: Single Node Trees

If both trees are single nodes with the same value, they are mirrors. If either node has children and the other doesn’t, they’re not mirrors.

Finally

This problem teaches the power of recursion and symmetry. The key idea is to mirror the traversal — compare left of one with right of the other and continue down the tree. Always consider the base cases carefully, and remember to match both structure and value. With these principles, you can confidently identify mirror trees!

Algorithm Steps❯

Given two binary trees, if both trees are empty, they are mirrors; if one is empty and the other is not, they are not mirrors.
Compare the root nodes of both trees. If the values are not equal, the trees are not mirrors.
Recursively check if the left subtree of the first tree is a mirror of the right subtree of the second tree, and if the right subtree of the first tree is a mirror of the left subtree of the second tree.
If both recursive checks return true, then the two trees are mirror images of each other.

Code

Python

Java

JavaScript

C++

Kotlin

Swift

Php

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def isMirror(t1, t2):
    if not t1 and not t2:
        return True
    if not t1 or not t2:
        return False
    return (t1.val == t2.val) and isMirror(t1.left, t2.right) and isMirror(t1.right, t2.left)

# Example usage:
if __name__ == '__main__':
    # Tree 1:
    #      1
    #     / \
    #    2   3
    #
    # Tree 2 (mirror of Tree 1):
    #      1
    #     / \
    #    3   2
    tree1 = TreeNode(1, TreeNode(2), TreeNode(3))
    tree2 = TreeNode(1, TreeNode(3), TreeNode(2))
    print(isMirror(tree1, tree2))

⬅ Previous TopicCheck if a Binary Tree Contains Duplicate Subtrees of Size 2 or More

Next Topic ⮕Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree

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