Symmetric Tree (Mirror Image of itself)

Difficulty Level : Medium
Last Updated : 01 Apr, 2021

Given a binary tree, check whether it is a mirror of itself.

For example, this binary tree is symmetric:

     1
   /   \
  2     2
 / \   / \
3   4 4   3

But the following is not:
    1
   / \
  2   2
   \   \
   3    3

The idea is to write a recursive function isMirror() that takes two trees as an argument and returns true if trees are the mirror and false if trees are not mirrored. The isMirror() function recursively checks two roots and subtrees under the root.

Below is the implementation of the above algorithm.

C++14

// C++ program to check if a given
// Binary Tree is symmetric or not
#include <bits/stdc++.h>
using namespace std;
 
// A Binary Tree Node
struct Node
{
    int key;
    struct Node *left, *right;
};
 
// Utility function to create new Node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return (temp);
}
 
// Returns true if trees
// with roots as root1 and root2 are mirror
bool isMirror(struct Node* root1, struct Node* root2)
{
    // If both trees are emptu,
    // then they are mirror images
    if (root1 == NULL && root2 == NULL)
        return true;
 
    // For two trees to be mirror
    // images, the following
    // three conditions must be true
    // 1 - Their root node's
    // key must be same 2 - left
    // subtree of left tree and right subtree
    // of right tree have to be mirror images
    // 3 - right subtree of left tree and left subtree
    // of right tree have to be mirror images
    if (root1 && root2 && root1->key == root2->key)
        return isMirror(root1->left, root2->right)
               && isMirror(root1->right, root2->left);
 
    // if none of above conditions is true then root1
    // and root2 are not mirror images
    return false;
}
 
// Returns true if a tree is
// symmetric i.e. mirror image of itself
bool isSymmetric(struct Node* root)
{
    // Check if tree is mirror of itself
    return isMirror(root, root);
}
 
// Driver code
int main()
{
    // Let us construct the Tree shown in the above figure
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(2);
    root->left->left = newNode(3);
    root->left->right = newNode(4);
    root->right->left = newNode(4);
    root->right->right = newNode(3);
 
    if(isSymmetric(root))
      cout<<"Symmetric";
    else
      cout<<"Not symmetric";
    return 0;
}

Java

// Java program to check is
// binary tree is symmetric or not
class Node {
    int key;
    Node left, right;
 
    Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    Node root;
 
    // returns true if trees
    //  with roots as root1 and root2are mirror
    boolean isMirror(Node node1, Node node2)
    {
        // if both trees are empty,
        //  then they are mirror image
        if (node1 == null && node2 == null)
            return true;
 
        // For two trees to be mirror images, the following
        // three conditions must be true 1 - Their root
        // node's key must be same 2 - left subtree of left
        // tree and right subtree
        //      of right tree have to be mirror images
        // 3 - right subtree of left tree and left subtree
        //      of right tree have to be mirror images
        if (node1 != null && node2 != null
            && node1.key == node2.key)
            return (isMirror(node1.left, node2.right)
                    && isMirror(node1.right, node2.left));
 
        // if none of the above conditions is true then
        // root1 and root2 are not mirror images
        return false;
    }
 
    // returns true if the tree is symmetric i.e
    // mirror image of itself
    boolean isSymmetric()
    {
        // check if tree is mirror of itself
        return isMirror(root, root);
    }
 
    // Driver code
    public static void main(String args[])
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(2);
        tree.root.left.left = new Node(3);
        tree.root.left.right = new Node(4);
        tree.root.right.left = new Node(4);
        tree.root.right.right = new Node(3);
        boolean output = tree.isSymmetric();
        if (output == true)
            System.out.println("Symmetric");
        else
            System.out.println("Not symmetric");
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python

# Python program to check if a
# given Binary Tree is symmetric or not
 
# Node structure
 
 
class Node:
 
    # Utility function to create new node
    def __init__(self, key):
        self.key = key
        self.left = None
        self.right = None
 
# Returns True if trees
#with roots as root1 and root 2  are mirror
 
 
def isMirror(root1, root2):
    # If both trees are empty, then they are mirror images
    if root1 is None and root2 is None:
        return True
 
    """ For two trees to be mirror images,
        the following three conditions must be true
        1 - Their root node's key must be same
        2 - left subtree of left tree and right subtree
          of the right tree have to be mirror images
        3 - right subtree of left tree and left subtree
           of right tree have to be mirror images
    """
    if (root1 is not None and root2 is not None):
        if root1.key == root2.key:
            return (isMirror(root1.left, root2.right)and
                    isMirror(root1.right, root2.left))
 
    # If none of the above conditions is true then root1
    # and root2 are not mirror images
    return False
 
 
def isSymmetric(root):
 
    # Check if tree is mirror of itself
    return isMirror(root, root)
 
 
# Driver Code
# Let's construct the tree show in the above figure
root = Node(1)
root.left = Node(2)
root.right = Node(2)
root.left.left = Node(3)
root.left.right = Node(4)
root.right.left = Node(4)
root.right.right = Node(3)
print "Symmetric" if isSymmetric(root) == True else "Not symmetric"
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# program to check is binary
// tree is symmetric or not
using System;
 
class Node {
    public int key;
    public Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class GFG {
    Node root;
 
    // returns true if trees with roots
    // as root1 and root2 are mirror
    Boolean isMirror(Node node1, Node node2)
    {
        // if both trees are empty,
        // then they are mirror image
        if (node1 == null && node2 == null)
            return true;
 
        // For two trees to be mirror images,
        // the following three conditions must be true
        // 1 - Their root node's key must be same
        // 2 - left subtree of left tree and right
        // subtree of right tree have to be mirror images
        // 3 - right subtree of left tree and left subtree
        // of right tree have to be mirror images
        if (node1 != null && node2 != null
            && node1.key == node2.key)
            return (isMirror(node1.left, node2.right)
                    && isMirror(node1.right, node2.left));
 
        // if none of the above conditions
        // is true then root1 and root2 are
        // mirror images
        return false;
    }
 
    // returns true if the tree is symmetric
    // i.e mirror image of itself
    Boolean isSymmetric()
    {
        // check if tree is mirror of itself
        return isMirror(root, root);
    }
 
    // Driver Code
    static public void Main(String[] args)
    {
        GFG tree = new GFG();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(2);
        tree.root.left.left = new Node(3);
        tree.root.left.right = new Node(4);
        tree.root.right.left = new Node(4);
        tree.root.right.right = new Node(3);
        Boolean output = tree.isSymmetric();
        if (output == true)
            Console.WriteLine("Symmetric");
        else
            Console.WriteLine("Not symmetric");
    }
}
 
// This code is contributed by Arnab Kundu

Output

Symmetric

Time Complexity: O(N)

Auxiliary Space: O(h) where h is the maximum height of the tree

?list=PLqM7alHXFySHCXD7r1J0ky9Zg_GBB1dbk
Check for Symmetric Binary Tree (Iterative Approach)
This article is contributed by Muneer Ahmed. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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