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Given a binary tree, the task is to create a new binary tree which is a mirror image of the given binary tree.

Examples: 

Input:
        5
       / \
      3   6
     / \
    2   4
Output:
Inorder of original tree: 2 3 4 5 6 
Inorder of mirror tree: 6 5 4 3 2
Mirror tree will be:
  5
 / \
6   3
   / \
  4   2

Input:
        2
       / \
      1   8
     /     \
    12      9
Output:
Inorder of original tree: 12 1 2 8 9 
Inorder of mirror tree: 9 8 2 1 12

Approach: Write a recursive function that will take two nodes as the argument, one of the original tree and the other of the newly created tree. Now, for every passed node of the original tree, create a corresponding node in the mirror tree and then recursively call the same method for the child nodes but passing the left child of the original tree node with the right child of the mirror tree node and the right child of the original tree node with the left child of the mirror tree node.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// A binary tree node has data, pointer to
// left child and a pointer to right child
typedef struct treenode {
    int val;
    struct treenode* left;
    struct treenode* right;
} node;
 
// Helper function that allocates a new node with the
// given data and NULL left and right pointers
node* createNode(int val)
{
    node* newNode = (node*)malloc(sizeof(node));
    newNode->val = val;
    newNode->left = NULL;
    newNode->right = NULL;
    return newNode;
}
 
// Helper function to print Inorder traversal
void inorder(node* root)
{
    if (root == NULL)
        return;
    inorder(root->left);
    cout <<" "<< root->val;
    inorder(root->right);
}
 
// mirrorify function takes two trees,
// original tree and a mirror tree
// It recurses on both the trees,
// but when original tree recurses on left,
// mirror tree recurses on right and
// vice-versa
void mirrorify(node* root, node** mirror)
{
    if (root == NULL) {
        mirror = NULL;
        return;
    }
 
    // Create new mirror node from original tree node
    *mirror = createNode(root->val);
    mirrorify(root->left, &((*mirror)->right));
    mirrorify(root->right, &((*mirror)->left));
}
 
// Driver code
int main()
{
 
    node* tree = createNode(5);
    tree->left = createNode(3);
    tree->right = createNode(6);
    tree->left->left = createNode(2);
    tree->left->right = createNode(4);
 
    // Print inorder traversal of the input tree
    cout <<"Inorder of original tree: ";
    inorder(tree);
    node* mirror = NULL;
    mirrorify(tree, &mirror);
 
    // Print inorder traversal of the mirror tree
    cout <<"\nInorder of mirror tree: ";
    inorder(mirror);
 
    return 0;
}
 
// This code is contributed by shivanisinghss2110


C




// C implementation of the approach
#include <malloc.h>
#include <stdio.h>
 
// A binary tree node has data, pointer to
// left child and a pointer to right child
typedef struct treenode {
    int val;
    struct treenode* left;
    struct treenode* right;
} node;
 
// Helper function that allocates a new node with the
// given data and NULL left and right pointers
node* createNode(int val)
{
    node* newNode = (node*)malloc(sizeof(node));
    newNode->val = val;
    newNode->left = NULL;
    newNode->right = NULL;
    return newNode;
}
 
// Helper function to print Inorder traversal
void inorder(node* root)
{
    if (root == NULL)
        return;
    inorder(root->left);
    printf("%d ", root->val);
    inorder(root->right);
}
 
// mirrorify function takes two trees,
// original tree and a mirror tree
// It recurses on both the trees,
// but when original tree recurses on left,
// mirror tree recurses on right and
// vice-versa
void mirrorify(node* root, node** mirror)
{
    if (root == NULL) {
        mirror = NULL;
        return;
    }
 
    // Create new mirror node from original tree node
    *mirror = createNode(root->val);
    mirrorify(root->left, &((*mirror)->right));
    mirrorify(root->right, &((*mirror)->left));
}
 
// Driver code
int main()
{
 
    node* tree = createNode(5);
    tree->left = createNode(3);
    tree->right = createNode(6);
    tree->left->left = createNode(2);
    tree->left->right = createNode(4);
 
    // Print inorder traversal of the input tree
    printf("Inorder of original tree: ");
    inorder(tree);
    node* mirror = NULL;
    mirrorify(tree, &mirror);
 
    // Print inorder traversal of the mirror tree
    printf("\nInorder of mirror tree: ");
    inorder(mirror);
 
    return 0;
}


Java




// Java implementation of the approach
import java.util.Comparator;
 
class GFG
{
 
// A binary tree node has data, pointer to
// left child and a pointer to right child
static class node
{
    int val;
    node left;
    node right;
}
 
// Helper function that allocates
// a new node with the given data
// and null left and right pointers
static node createNode(int val)
{
    node newNode = new node();
    newNode.val = val;
    newNode.left = null;
    newNode.right = null;
    return newNode;
}
 
// Helper function to print Inorder traversal
static void inorder(node root)
{
    if (root == null)
        return;
    inorder(root.left);
    System.out.print(root.val);
    inorder(root.right);
}
 
// mirrorify function takes two trees,
// original tree and a mirror tree
// It recurses on both the trees,
// but when original tree recurses on left,
// mirror tree recurses on right and
// vice-versa
static node mirrorify(node root)
{
    if (root == null)
    {
        return null;
         
    }
 
    // Create new mirror node from original tree node
    node mirror = createNode(root.val);
    mirror.right = mirrorify(root.left);
    mirror.left = mirrorify(root.right);
    return mirror;
}
 
// Driver code
public static void main(String args[])
{
 
    node tree = createNode(5);
    tree.left = createNode(3);
    tree.right = createNode(6);
    tree.left.left = createNode(2);
    tree.left.right = createNode(4);
 
    // Print inorder traversal of the input tree
    System.out.print("Inorder of original tree: ");
    inorder(tree);
    node mirror = null;
    mirror = mirrorify(tree);
 
    // Print inorder traversal of the mirror tree
    System.out.print("\nInorder of mirror tree: ");
    inorder(mirror);
}
}
 
// This code is contributed by Arnab Kundu


Python3




# Python3 implementation of the approach
 
# A binary tree node has data,
# pointer to left child and
# a pointer to right child
# Linked list node
class Node:
    def __init__(self, data):
        self.data = data
        self.left = None
        self.right = None
 
# Helper function that allocates
# a new node with the given data
# and None left and right pointers
def createNode(val):
    newNode = Node(0)
    newNode.val = val
    newNode.left = None
    newNode.right = None
    return newNode
 
# Helper function to print Inorder traversal
def inorder(root):
    if (root == None):
        return
    inorder(root.left)
    print( root.val, end = " ")
    inorder(root.right)
 
# mirrorify function takes two trees,
# original tree and a mirror tree
# It recurses on both the trees,
# but when original tree recurses on left,
# mirror tree recurses on right and
# vice-versa
def mirrorify(root, mirror):
 
    if (root == None) :
        mirror = None
        return mirror
     
    # Create new mirror node
    # from original tree node
    mirror = createNode(root.val)
    mirror.right = mirrorify(root.left,
                           ((mirror).right))
    mirror.left = mirrorify(root.right,
                          ((mirror).left))
    return mirror
 
# Driver Code
if __name__=='__main__':
 
    tree = createNode(5)
    tree.left = createNode(3)
    tree.right = createNode(6)
    tree.left.left = createNode(2)
    tree.left.right = createNode(4)
 
    # Print inorder traversal of the input tree
    print("Inorder of original tree: ")
    inorder(tree)
    mirror = None
    mirror = mirrorify(tree, mirror)
 
    # Print inorder traversal of the mirror tree
    print("\nInorder of mirror tree: ")
    inorder(mirror)
 
# This code is contributed by Arnab Kundu


C#




// c# implementation of the approach
using System;
public class GFG
{
 
// A binary tree node has data, pointer to
// left child and a pointer to right child
public class node
{
    public int val;
    public node left;
    public node right;
}
 
// Helper function that allocates
// a new node with the given data
// and null left and right pointers
public static node createNode(int val)
{
    node newNode = new node();
    newNode.val = val;
    newNode.left = null;
    newNode.right = null;
    return newNode;
}
 
// Helper function to print Inorder traversal
public static void inorder(node root)
{
    if (root == null)
    {
        return;
    }
    inorder(root.left);
    Console.Write("{0:D} ", root.val);
    inorder(root.right);
}
 
// mirrorify function takes two trees,
// original tree and a mirror tree
// It recurses on both the trees,
// but when original tree recurses on left,
// mirror tree recurses on right and
// vice-versa
public static node mirrorify(node root)
{
    if (root == null)
    {
        return null;
 
    }
 
    // Create new mirror node from original tree node
    node mirror = createNode(root.val);
    mirror.right = mirrorify(root.left);
    mirror.left = mirrorify(root.right);
    return mirror;
}
 
// Driver code
public static void Main(string[] args)
{
 
    node tree = createNode(5);
    tree.left = createNode(3);
    tree.right = createNode(6);
    tree.left.left = createNode(2);
    tree.left.right = createNode(4);
 
    // Print inorder traversal of the input tree
    Console.Write("Inorder of original tree: ");
    inorder(tree);
    node mirror = null;
    mirror = mirrorify(tree);
 
    // Print inorder traversal of the mirror tree
    Console.Write("\nInorder of mirror tree: ");
    inorder(mirror);
}
}


Javascript




<script>
// Javascript implementation of the approach
 
// A binary tree node has data, pointer to
// left child and a pointer to right child
class node
{
    constructor()
    {
        this.val=0;
        this.left=this.right=null;
    }
}
 
// Helper function that allocates
// a new node with the given data
// and null left and right pointers
function createNode(val)
{
    let newNode = new node();
    newNode.val = val;
    newNode.left = null;
    newNode.right = null;
    return newNode;
}
 
// Helper function to print Inorder traversal
function inorder(root)
{
    if (root == null)
        return;
    inorder(root.left);
    document.write(root.val+" ");
    inorder(root.right);
}
 
// mirrorify function takes two trees,
// original tree and a mirror tree
// It recurses on both the trees,
// but when original tree recurses on left,
// mirror tree recurses on right and
// vice-versa
function mirrorify(root)
{
    if (root == null)
    {
        return null;
          
    }
  
    // Create new mirror node from original tree node
    let mirror = createNode(root.val);
    mirror.right = mirrorify(root.left);
    mirror.left = mirrorify(root.right);
    return mirror;
}
 
// Driver code
let tree = createNode(5);
tree.left = createNode(3);
tree.right = createNode(6);
tree.left.left = createNode(2);
tree.left.right = createNode(4);
 
// Print inorder traversal of the input tree
document.write("Inorder of original tree: ");
inorder(tree);
let mirror = null;
mirror = mirrorify(tree);
 
// Print inorder traversal of the mirror tree
document.write("<br>Inorder of mirror tree: ");
inorder(mirror);
 
// This code is contributed by avanitrachhadiya2155
</script>


Output

Inorder of original tree: 2 3 4 5 6 
Inorder of mirror tree: 6 5 4 3 2 

Time Complexity: O(n), where n is the number of nodes in the tree. This is because we need to visit each node in the tree exactly once to swap its left and right child nodes.
Auxiliary Space: O(h), where h is the height of the binary tree. This is because the maximum amount of space used by the algorithm at any given time is the size of the call stack, which is at most equal to the height of the binary tree. This is because each recursive call to mirror adds a new frame to the call stack, and the stack grows to a maximum depth of the height of the tree. Therefore, the space used by the algorithm is proportional to the height of the tree.

Approach 2:
 In order to change the original tree in its mirror tree, then we simply swap the left and right link of each node. If the node is leaf node then do nothing.

C++




#include <iostream>
using namespace std;
 
typedef struct treenode {
    int val;
    struct treenode* left;
    struct treenode* right;
} node;
 
// Helper function that
// allocates a new node with the
// given data and NULL left and right pointers
node* createNode(int val)
{
    node* newNode = (node*)malloc(sizeof(node));
    newNode->val = val;
    newNode->left = NULL;
    newNode->right = NULL;
    return newNode;
}
 
// Function to print the inrder traversal
void inorder(node* root)
{
    if (root == NULL)
        return;
    inorder(root->left);
    printf("%d ", root->val);
    inorder(root->right);
}
 
// Function to convert to  mirror tree
treenode* mirrorTree(node* root)
{
    // Base Case
    if (root == NULL)
        return root;
    node* t = root->left;
    root->left = root->right;
    root->right = t;
 
    if (root->left)
        mirrorTree(root->left);
    if (root->right)
        mirrorTree(root->right);
   
  return root;
}
 
// Driver Code
int main()
{
 
    node* tree = createNode(5);
    tree->left = createNode(3);
    tree->right = createNode(6);
    tree->left->left = createNode(2);
    tree->left->right = createNode(4);
    printf("Inorder of original tree: ");
    inorder(tree);
 
    // Function call
    mirrorTree(tree);
 
    printf("\nInorder of Mirror tree: ");
    inorder(tree);
    return 0;
}


Java




import java.util.*;
 
class GFG{
     
static class Node
{
    int val;
    Node left;
    Node right;
}
 
// Helper function that allocates
// a new node with the given data
// and null left and right references
public static Node createNode(int val)
{
    Node newNode = new Node();
    newNode.val = val;
    newNode.left = null;
    newNode.right = null;
    return newNode;
}
 
// Function to print the inorder traversal
public static void inOrder(Node root)
{
    if (root == null)
        return;
 
    inOrder(root.left);
    System.out.print(root.val + " ");
    inOrder(root.right);
}
 
// Function to convert to mirror tree
// by swapping the left and right links.
public static Node mirrorTree(Node root)
{
    if (root == null)
        return null;
 
    Node left = mirrorTree(root.left);
    Node right = mirrorTree(root.right);
 
    root.left = right;
    root.right = left;
 
    return root;
}
 
// Driver Code
public static void main(String args[])
{
    Node tree = createNode(5);
    tree.left = createNode(3);
    tree.right = createNode(6);
    tree.left.left = createNode(2);
    tree.left.right = createNode(4);
    System.out.print("Inorder of original tree: ");
    inOrder(tree);
 
    // Function call
    mirrorTree(tree);
 
    System.out.print("\nInorder of mirror tree: ");
    inOrder(tree);
}
}
 
// This code is contributed by Ryan Ranaut


Python3




# code
class Node:
   def __init__(self,data):
       self.left = None
       self.right = None
       self.data = data
 
def inOrder(root):
   if root is not None:
       inOrder(root.left)
       print (root.data, end = ' ')
       inOrder(root.right)
 
#we recursively call the mirror function which swaps the right subtree with the left subtree.
def mirror(root):
    if root is None:
        return
    mirror(root.left)
    mirror(root.right)
    root.left, root.right = root.right, root.left
 
 
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.right.left = Node(5)
 
print("The inorder traversal of the tree before mirroring:",end=' ')
print(inOrder(root))
# 4 2 1 5 3
print()
mirror(root)
print("The inorder traversal of the tree after mirroring:",end=' ')
print(inOrder(root))
# 3 5 1 2 4


C#




using System;
 
public class GFG{
     
public class Node
{
    public int val;
    public Node left;
    public Node right;
}
 
// Helper function that allocates
// a new node with the given data
// and null left and right references
public static Node createNode(int val)
{
    Node newNode = new Node();
    newNode.val = val;
    newNode.left = null;
    newNode.right = null;
    return newNode;
}
 
// Function to print the inorder traversal
public static void inOrder(Node root)
{
    if (root == null)
        return;
 
    inOrder(root.left);
    Console.Write(root.val + " ");
    inOrder(root.right);
}
 
// Function to convert to mirror tree
// by swapping the left and right links.
public static Node mirrorTree(Node root)
{
    if (root == null)
        return null;
 
    Node left = mirrorTree(root.left);
    Node right = mirrorTree(root.right);
 
    root.left = right;
    root.right = left;
 
    return root;
}
 
// Driver Code
public static void Main(String []args)
{
    Node tree = createNode(5);
    tree.left = createNode(3);
    tree.right = createNode(6);
    tree.left.left = createNode(2);
    tree.left.right = createNode(4);
    Console.Write("Inorder of original tree: ");
    inOrder(tree);
 
    // Function call
    mirrorTree(tree);
 
    Console.Write("\nInorder of mirror tree: ");
    inOrder(tree);
}
}
 
// This code is contributed by shivanisinghss2110


Javascript




<script>
 
     
class Node
{
    constructor()
    {
        this.val=0;
        this.left=this.right=null;
    }
}
 
// Helper function that allocates
// a new node with the given data
// and null left and right references
function createNode(val)
{
    let newNode = new Node();
    newNode.val = val;
    newNode.left = null;
    newNode.right = null;
    return newNode;
}
 
// Function to print the inorder traversal
function inOrder(root)
{
    if (root == null)
        return;
 
    inOrder(root.left);
    document.write(root.val + " ");
    inOrder(root.right);
}
 
// Function to convert to mirror tree
// by swapping the left and right links.
function mirrorTree(root)
{
    if (root == null)
        return null;
 
    let left = mirrorTree(root.left);
    let right = mirrorTree(root.right);
 
    root.left = right;
    root.right = left;
 
    return root;
}
 
// Driver Code
    let tree = createNode(5);
    tree.left = createNode(3);
    tree.right = createNode(6);
    tree.left.left = createNode(2);
    tree.left.right = createNode(4);
    document.write("Inorder of original tree: " );
    inOrder(tree);
 
    // Function call
    mirrorTree(tree);
 
    document.write("<br>" +"\nInorder of mirror tree: ");
    inOrder(tree);
 
 
// This code is contributed by shivanisinghss2110
</script>


Output

Inorder of original tree: 2 3 4 5 6 
Inorder of Mirror tree: 6 5 4 3 2 

Time complexity: O(N) where N is the number of nodes in given binary tree.
Auxiliary Space: O(h) where h is the height of the tree. due to recursion call stack.



Last Updated : 26 Mar, 2023
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