Convert a Binary Tree into its Mirror Tree

• Difficulty Level : Easy
• Last Updated : 10 Jul, 2021

Mirror of a Tree: Mirror of a Binary Tree T is another Binary Tree M(T) with left and right children of all non-leaf nodes interchanged. Trees in the above figure are mirror of each other

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Method 1 (Recursive)

Algorithm – Mirror(tree):

(1)  Call Mirror for left-subtree    i.e., Mirror(left-subtree)
(2)  Call Mirror for right-subtree  i.e., Mirror(right-subtree)
(3)  Swap left and right subtrees.
temp = left-subtree
left-subtree = right-subtree
right-subtree = temp

C++

// C++ program to convert a binary tree
// to its mirror
#include<bits/stdc++.h>
using namespace std;

/* A binary tree node has data, pointer
to left child and a pointer to right child */
struct Node
{
int data;
struct Node* left;
struct Node* right;
};

/* Helper function that allocates a new node with
the given data and NULL left and right pointers. */
struct Node* newNode(int data)
{
struct Node* node = (struct Node*)
malloc(sizeof(struct Node));
node->data = data;
node->left = NULL;
node->right = NULL;

return(node);
}

/* Change a tree so that the roles of the left and
right pointers are swapped at every node.

So the tree...
4
/ \
2 5
/ \
1 3

is changed to...
4
/ \
5 2
/ \
3 1
*/
void mirror(struct Node* node)
{
if (node == NULL)
return;
else
{
struct Node* temp;

/* do the subtrees */
mirror(node->left);
mirror(node->right);

/* swap the pointers in this node */
temp     = node->left;
node->left = node->right;
node->right = temp;
}
}

/* Helper function to print
Inorder traversal.*/
void inOrder(struct Node* node)
{
if (node == NULL)
return;

inOrder(node->left);
cout << node->data << " ";
inOrder(node->right);
}

// Driver Code
int main()
{
struct Node *root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);

/* Print inorder traversal of the input tree */
cout << "Inorder traversal of the constructed"
<< " tree is" << endl;
inOrder(root);

/* Convert tree to its mirror */
mirror(root);

/* Print inorder traversal of the mirror tree */
cout << "\nInorder traversal of the mirror tree"
<< " is \n";
inOrder(root);

return 0;
}

// This code is contributed by Akanksha Rai

C

// C program to convert a binary tree
// to its mirror
#include<stdio.h>
#include<stdlib.h>

/* A binary tree node has data, pointer
to left child and a pointer to right child */
struct Node
{
int data;
struct Node* left;
struct Node* right;
};

/* Helper function that allocates a new node with the
given data and NULL left and right pointers. */
struct Node* newNode(int data)

{
struct Node* node = (struct Node*)
malloc(sizeof(struct Node));
node->data = data;
node->left = NULL;
node->right = NULL;

return(node);
}

/* Change a tree so that the roles of the  left and
right pointers are swapped at every node.

So the tree...
4
/ \
2   5
/ \
1   3

is changed to...
4
/ \
5   2
/ \
3   1
*/
void mirror(struct Node* node)
{
if (node==NULL)
return;
else
{
struct Node* temp;

/* do the subtrees */
mirror(node->left);
mirror(node->right);

/* swap the pointers in this node */
temp        = node->left;
node->left  = node->right;
node->right = temp;
}
}

/* Helper function to print Inorder traversal.*/
void inOrder(struct Node* node)
{
if (node == NULL)
return;

inOrder(node->left);
printf("%d ", node->data);
inOrder(node->right);
}

/* Driver program to test mirror() */
int main()
{
struct Node *root = newNode(1);
root->left        = newNode(2);
root->right       = newNode(3);
root->left->left  = newNode(4);
root->left->right = newNode(5);

/* Print inorder traversal of the input tree */
printf("Inorder traversal of the constructed"
" tree is \n");
inOrder(root);

/* Convert tree to its mirror */
mirror(root);

/* Print inorder traversal of the mirror tree */
printf("\nInorder traversal of the mirror tree"
" is \n");
inOrder(root);

return 0;
}

Java

// Java program to convert binary tree into its mirror

/* Class containing left and right child of current
node and key value*/
class Node
{
int data;
Node left, right;

public Node(int item)
{
data = item;
left = right = null;
}
}

class BinaryTree
{
Node root;

void mirror()
{
root = mirror(root);
}

Node mirror(Node node)
{
if (node == null)
return node;

/* do the subtrees */
Node left = mirror(node.left);
Node right = mirror(node.right);

/* swap the left and right pointers */
node.left = right;
node.right = left;

return node;
}

void inOrder()
{
inOrder(root);
}

/* Helper function to test mirror(). Given a binary
search tree, print out its data elements in
increasing sorted order.*/
void inOrder(Node node)
{
if (node == null)
return;

inOrder(node.left);
System.out.print(node.data + " ");

inOrder(node.right);
}

/* testing for example nodes */
public static void main(String args[])
{
/* creating a binary tree and entering the nodes */
BinaryTree tree = new BinaryTree();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);

/* print inorder traversal of the input tree */
System.out.println("Inorder traversal of input tree is :");
tree.inOrder();
System.out.println("");

/* convert tree to its mirror */
tree.mirror();

/* print inorder traversal of the minor tree */
System.out.println("Inorder traversal of binary tree is : ");
tree.inOrder();

}
}

Python3

# Python3 program to convert a binary
# tree to its mirror

# Utility function to create a new
# tree node
class newNode:
def __init__(self,data):
self.data = data
self.left = self.right = None

""" Change a tree so that the roles of the
left and right pointers are swapped at
every node.

So the tree...
4
/ \
2 5
/ \
1 3

is changed to...
4
/ \
5 2
/ \
3 1
"""
def mirror(node):

if (node == None):
return
else:

temp = node

""" do the subtrees """
mirror(node.left)
mirror(node.right)

""" swap the pointers in this node """
temp = node.left
node.left = node.right
node.right = temp

""" Helper function to print Inorder traversal."""
def inOrder(node) :

if (node == None):
return

inOrder(node.left)
print(node.data, end = " ")
inOrder(node.right)

# Driver code
if __name__ =="__main__":

root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.left.right = newNode(5)

""" Print inorder traversal of
the input tree """
print("Inorder traversal of the",
"constructed tree is")
inOrder(root)

""" Convert tree to its mirror """
mirror(root)

""" Print inorder traversal of
the mirror tree """
print("\nInorder traversal of",
"the mirror treeis ")
inOrder(root)

# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

C#

// C# program to convert binary
// tree into its mirror
using System;

// Class containing left and right
// child of current node and key value
public class Node
{
public int data;
public Node left, right;

public Node(int item)
{
data = item;
left = right = null;
}
}

class GFG
{
public Node root;

public virtual void mirror()
{
root = mirror(root);
}

public virtual Node mirror(Node node)
{
if (node == null)
{
return node;
}

/* do the subtrees */
Node left = mirror(node.left);
Node right = mirror(node.right);

/* swap the left and right pointers */
node.left = right;
node.right = left;

return node;
}

public virtual void inOrder()
{
inOrder(root);
}

/* Helper function to test mirror().
Given a binary search tree, print out its
data elements in increasing sorted order.*/
public virtual void inOrder(Node node)
{
if (node == null)
{
return;
}

inOrder(node.left);
Console.Write(node.data + " ");

inOrder(node.right);
}

/* testing for example nodes */
public static void Main(string[] args)
{
/* creating a binary tree and
entering the nodes */
GFG tree = new GFG();
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);

/* print inorder traversal of the input tree */
Console.WriteLine("Inorder traversal " +
"of input tree is :");
tree.inOrder();
Console.WriteLine("");

/* convert tree to its mirror */
tree.mirror();

/* print inorder traversal of the minor tree */
Console.WriteLine("Inorder traversal " +
"of binary tree is : ");
tree.inOrder();
}
}

// This code is contributed by Shrikant13

Javascript

<script>

// JavaScript program to convert
// binary tree into its mirror

/* Class containing left and right child of current
node and key value*/

class Node
{
constructor(item)
{
this.data=item;
this.left=this.right=null;
}
}

let root;

function mirror(node)
{
if (node == null)
return node;

/* do the subtrees */
let left = mirror(node.left);
let right = mirror(node.right);

/* swap the left and right pointers */
node.left = right;
node.right = left;

return node;
}

/* Helper function to test mirror(). Given a binary
search tree, print out its data elements in
increasing sorted order.*/
function inOrder(node)
{
if (node == null)
return;

inOrder(node.left);
document.write(node.data + " ");

inOrder(node.right);
}

/* testing for example nodes */
/* creating a binary tree and entering the nodes */

root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);

/* print inorder traversal of the input tree */
document.write("Inorder traversal of input tree is :<br>");
inOrder(root);
document.write("<br>");

/* convert tree to its mirror */
mirror(root);

/* print inorder traversal of the minor tree */
document.write(
"Inorder traversal of binary tree is : <br>"
);
inOrder(root);

// This code is contributed by rag2127

</script>

Output:

Inorder traversal of the constructed tree is
4 2 5 1 3
Inorder traversal of the mirror tree is
3 1 5 2 4

Time & Space Complexities: Worst-case Time complexity is O(n) and for space complexity, If we don’t consider the size of the recursive stack for function calls then O(1) otherwise O(h) where h is the height of the tree.  This program is similar to traversal of tree space and time complexities will be the same as Tree traversal more for information Please see our Tree Traversal post for details.

Method 2 (Iterative)

The idea is to do queue based level order traversal. While doing traversal, swap left and right children of every node.

C++

// Iterative CPP program to convert a Binary
// Tree to its mirror
#include<bits/stdc++.h>
using namespace std;

/* A binary tree node has data, pointer to
left child and a pointer to right child */
struct Node
{
int data;
struct Node* left;
struct Node* right;
};

/* Helper function that allocates a new node
with the given data and NULL left and right
pointers. */
struct Node* newNode(int data)

{
struct Node* node = new Node;
node->data = data;
node->left = node->right = NULL;
return(node);
}

/* Change a tree so that the roles of the  left and
right pointers are swapped at every node.
So the tree...
4
/ \
2   5
/ \
1   3

is changed to...
4
/ \
5   2
/ \
3   1
*/
void mirror(Node* root)
{
if (root == NULL)
return;

queue<Node*> q;
q.push(root);

// Do BFS. While doing BFS, keep swapping
// left and right children
while (!q.empty())
{
// pop top node from queue
Node* curr = q.front();
q.pop();

// swap left child with right child
swap(curr->left, curr->right);

// push left and right children
if (curr->left)
q.push(curr->left);
if (curr->right)
q.push(curr->right);
}
}

/* Helper function to print Inorder traversal.*/
void inOrder(struct Node* node)
{
if (node == NULL)
return;
inOrder(node->left);
cout << node->data << " ";
inOrder(node->right);
}

/* Driver program to test mirror() */
int main()
{
struct Node *root = newNode(1);
root->left        = newNode(2);
root->right       = newNode(3);
root->left->left  = newNode(4);
root->left->right = newNode(5);

/* Print inorder traversal of the input tree */
cout << "\n Inorder traversal of the"
" constructed tree is \n";
inOrder(root);

/* Convert tree to its mirror */
mirror(root);

/* Print inorder traversal of the mirror tree */
cout << "\n Inorder traversal of the "
"mirror tree is \n";
inOrder(root);

return 0;
}

Java

// Iterative Java program to convert a Binary
// Tree to its mirror
import java.util.*;

class GFG
{

/* A binary tree node has data, pointer to
left child and a pointer to right child */
static class Node
{
int data;
Node left;
Node right;
};

/* Helper function that allocates a new node
with the given data and null left and right
pointers. */
static Node newNode(int data)

{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return(node);
}

/* Change a tree so that the roles of the left and
right pointers are swapped at every node.
So the tree...
4
/ \
2 5
/ \
1 3

is changed to...
4
/ \
5 2
/ \
3 1
*/
static void mirror(Node root)
{
if (root == null)
return;

// Do BFS. While doing BFS, keep swapping
// left and right children
while (q.size() > 0)
{
// pop top node from queue
Node curr = q.peek();
q.remove();

// swap left child with right child
Node temp = curr.left;
curr.left = curr.right;
curr.right = temp;;

// push left and right children
if (curr.left != null)
if (curr.right != null)
}
}

/* Helper function to print Inorder traversal.*/
static void inOrder( Node node)
{
if (node == null)
return;
inOrder(node.left);
System.out.print( node.data + " ");
inOrder(node.right);
}

/* Driver code */
public static void main(String args[])
{
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);

/* Print inorder traversal of the input tree */
System.out.print( "\n Inorder traversal of the"
+" coned tree is \n");
inOrder(root);

/* Convert tree to its mirror */
mirror(root);

/* Print inorder traversal of the mirror tree */
System.out.print( "\n Inorder traversal of the "+
"mirror tree is \n");
inOrder(root);
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to convert a Binary
# Tree to its mirror

# A binary tree node has data, pointer to
# left child and a pointer to right child
# Helper function that allocates a new node
# with the given data and None left and
# right pointers
class newNode:
def __init__(self, data):
self.data = data
self.left = None
self.right = None

''' Change a tree so that the roles of the left
and right pointers are swapped at every node.
So the tree...
4
/ \
2 5
/ \
1 3

is changed to...
4
/ \
5 2
/ \
3 1
'''

def mirror( root):

if (root == None):
return

q = []
q.append(root)

# Do BFS. While doing BFS, keep swapping
# left and right children
while (len(q)):

# pop top node from queue
curr = q
q.pop(0)

# swap left child with right child
curr.left, curr.right = curr.right, curr.left

# append left and right children
if (curr.left):
q.append(curr.left)
if (curr.right):
q.append(curr.right)

""" Helper function to print Inorder traversal."""
def inOrder( node):
if (node == None):
return
inOrder(node.left)
print(node.data, end = " ")
inOrder(node.right)

# Driver code
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.left.right = newNode(5)

""" Print inorder traversal of the input tree """
print("Inorder traversal of the constructed tree is")
inOrder(root)

""" Convert tree to its mirror """
mirror(root)

""" Print inorder traversal of the mirror tree """
print("\nInorder traversal of the mirror tree is")
inOrder(root)

# This code is contributed by SHUBHAMSINGH10

C#

// C# Iterative Java program to convert a Binary
// Tree to its mirror
using System.Collections.Generic;
using System;

class GFG
{

/* A binary tree node has data, pointer to
left child and a pointer to right child */
public class Node
{
public int data;
public Node left;
public Node right;
};

/* Helper function that allocates a new node
with the given data and null left and right
pointers. */
static Node newNode(int data)

{
Node node = new Node();
node.data = data;
node.left = node.right = null;
return(node);
}

/* Change a tree so that the roles of the left and
right pointers are swapped at every node.
So the tree...
4
/ \
2 5
/ \
1 3

is changed to...
4
/ \
5 2
/ \
3 1
*/
static void mirror(Node root)
{
if (root == null)
return;

Queue<Node> q = new Queue<Node>();
q.Enqueue(root);

// Do BFS. While doing BFS, keep swapping
// left and right children
while (q.Count > 0)
{
// pop top node from queue
Node curr = q.Peek();
q.Dequeue();

// swap left child with right child
Node temp = curr.left;
curr.left = curr.right;
curr.right = temp;;

// push left and right children
if (curr.left != null)
q.Enqueue(curr.left);
if (curr.right != null)
q.Enqueue(curr.right);
}
}

/* Helper function to print Inorder traversal.*/
static void inOrder( Node node)
{
if (node == null)
return;
inOrder(node.left);
Console.Write( node.data + " ");
inOrder(node.right);
}

/* Driver code */
public static void Main(String []args)
{
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);

/* Print inorder traversal of the input tree */
Console.Write( "\n Inorder traversal of the"
+" coned tree is \n");
inOrder(root);

/* Convert tree to its mirror */
mirror(root);

/* Print inorder traversal of the mirror tree */
Console.Write( "\n Inorder traversal of the "+
"mirror tree is \n");
inOrder(root);
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>
// Iterative Javascript program to convert a Binary
// Tree to its mirror

class Node
{
constructor(data) {
this.left = null;
this.right = null;
this.data = data;
}
}

/* Helper function that allocates a new node
with the given data and null left and right
pointers. */
function newNode(data)

{
let node = new Node(data);
return(node);
}

/* Change a tree so that the roles of the left and
right pointers are swapped at every node.
So the tree...
4
/ \
2 5
/ \
1 3

is changed to...
4
/ \
5 2
/ \
3 1
*/
function mirror(root)
{
if (root == null)
return;

let q = [];
q.push(root);

// Do BFS. While doing BFS, keep swapping
// left and right children
while (q.length > 0)
{
// pop top node from queue
let curr = q;
q.shift();

// swap left child with right child
let temp = curr.left;
curr.left = curr.right;
curr.right = temp;;

// push left and right children
if (curr.left != null)
q.push(curr.left);
if (curr.right != null)
q.push(curr.right);
}
}

/* Helper function to print Inorder traversal.*/
function inOrder(node)
{
if (node == null)
return;
inOrder(node.left);
document.write( node.data + " ");
inOrder(node.right);
}

let root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);

/* Print inorder traversal of the input tree */
document.write(" Inorder traversal of the"
+" constructed tree is " + "</br>");
inOrder(root);

/* Convert tree to its mirror */
mirror(root);

/* Print inorder traversal of the mirror tree */
document.write("</br>" + " Inorder traversal of the "+
"mirror tree is " + "</br>");
inOrder(root);

</script>

Output:

Inorder traversal of the constructed tree is
4 2 5 1 3
Inorder traversal of the mirror tree is
3 1 5 2 4

My Personal Notes arrow_drop_up
Recommended Articles
Page :