Create a mirror tree from the given binary tree

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

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// 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;
}

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Java

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// 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.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 
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.printf("Inorder of original tree: "); 
    inorder(tree); 
    node mirror = null
    mirror = mirrorify(tree); 
  
    // Print inorder traversal of the mirror tree 
    System.out.printf("\nInorder of mirror tree: "); 
    inorder(mirror); 
}
  
// This code is contributed by Arnab Kundu

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Python3

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# 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

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C#

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using System;
  
// c# implementation of the approach 
  
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);
}
}

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Output:

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


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