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Flip Binary Tree
  • Difficulty Level : Hard
  • Last Updated : 18 Nov, 2020

Given a binary tree, the task is to flip the binary tree towards the right direction that is clockwise. See the below examples to see the transformation.
In the flip operation, the leftmost node becomes the root of the flipped tree and its parent becomes its right child and the right sibling becomes its left child and the same should be done for all left most nodes recursively. 

tree1tree2

Below is the main rotation code of a subtree 

    root->left->left = root->right;
    root->left->right = root;
    root->left = NULL;
    root->right = NULL; 

The above code can be understood by the following diagram – 

tree4



as we are storing root->left in the flipped root, flipped subtree gets stored in each recursive call.

C++




/*  C/C++ program to flip a binary tree */
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node structure */
struct Node
{
    int data;
    Node *left, *right;
};
 
/* Utility function to create a new Binary
   Tree Node */
struct Node* newNode(int data)
{
    struct Node *temp = new struct Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// method to flip the binary tree
Node* flipBinaryTree(Node* root)
{
    // Base cases
    if (root == NULL)
        return root;
    if (root->left == NULL && root->right == NULL)
        return root;
 
    //  recursively call the same method
    Node* flippedRoot = flipBinaryTree(root->left);
 
    //  rearranging main root Node after returning
    // from recursive call
    root->left->left = root->right;
    root->left->right = root;
    root->left = root->right = NULL;
 
    return flippedRoot;
}
 
// Iterative method to do level order traversal
// line by line
void printLevelOrder(Node *root)
{
    // Base Case
    if (root == NULL)  return;
 
    // Create an empty queue for level order traversal
    queue<Node *> q;
 
    // Enqueue Root and initialize height
    q.push(root);
 
    while (1)
    {
        // nodeCount (queue size) indicates number
        // of nodes at current lelvel.
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
 
        // Dequeue all nodes of current level and
        // Enqueue all nodes of next level
        while (nodeCount > 0)
        {
            Node *node = q.front();
            cout << node->data << " ";
            q.pop();
            if (node->left != NULL)
                q.push(node->left);
            if (node->right != NULL)
                q.push(node->right);
            nodeCount--;
        }
        cout << endl;
    }
}
 
//  Driver code
int main()
{
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->right->left = newNode(4);
    root->right->right = newNode(5);
 
    cout << "Level order traversal of given tree\n";
    printLevelOrder(root);
 
    root = flipBinaryTree(root);
 
    cout << "\nLevel order traversal of the flipped"
            " tree\n";
    printLevelOrder(root);
    return 0;
}

Java




/*  Java program to flip a binary tree */
import java.util.Queue;
import java.util.LinkedList;
public class FlipTree {
 
    // method to flip the binary tree
    public static Node flipBinaryTree(Node root)
    {
        if (root == null)
            return root;
        if (root.left == null && root.right ==null)
            return root;
 
        //  recursively call the same method
        Node flippedRoot=flipBinaryTree(root.left);
 
        //  rearranging main root Node after returning
        // from recursive call
        root.left.left=root.right;
        root.left.right=root;
        root.left=root.right=null;
        return flippedRoot;
    }
 
    // Iterative method to do level order traversal
    // line by line
    public static void printLevelOrder(Node root)
    {
        // Base Case
        if(root==null)
            return ;
         
        // Create an empty queue for level order traversal
        Queue<Node> q=new LinkedList<>();
        // Enqueue Root and initialize height
        q.add(root);
        while(true)
        {
            // nodeCount (queue size) indicates number
            // of nodes at current lelvel.
            int nodeCount = q.size();
            if (nodeCount == 0)
                break;
             
            // Dequeue all nodes of current level and
            // Enqueue all nodes of next level
            while (nodeCount > 0)
            {
                Node node = q.remove();
                System.out.print(node.data+" ");
                if (node.left != null)
                    q.add(node.left);
                if (node.right != null)
                    q.add(node.right);
                nodeCount--;
            }
            System.out.println();
        }
    }
 
    public static void main(String args[]) {
        Node root=new Node(1);
        root.left=new Node(2);
        root.right=new Node(1);
        root.right.left = new Node(4);
        root.right.right = new Node(5);
        System.out.println("Level order traversal of given tree");
        printLevelOrder(root);
   
        root = flipBinaryTree(root);
        System.out.println("Level order traversal of flipped tree");
        printLevelOrder(root);
    }
}
 
/* A binary tree node structure */
class Node
{
    int data;
    Node left, right;
    Node(int data)
    {
        this.data=data;
    }
};
 
//This code is contributed by Gaurav Tiwari

Python3




# Python3 program to flip
# a binary tree
 
# A binary tree node
class Node:
     
    # Constructor to create
    # a new node
    def __init__(self, data):
       
        self.data = data
        self.right = None
        self.left = None
 
def flipBinaryTree(root):
     
    # Base Cases
    if root is None:
        return root
     
    if (root.left is None and
        root.right is None):
        return root
 
    # Recursively call the
    # same method
    flippedRoot = flipBinaryTree(root.left)
 
    # Rearranging main root Node
    # after returning from
    # recursive call
    root.left.left = root.right
    root.left.right = root
    root.left = root.right = None
 
    return flippedRoot
 
# Iterative method to do the level
# order traversal line by line
def printLevelOrder(root):
     
    # Base Case
    if root is None:
        return
     
    # Create an empty queue for
    # level order traversal
    from Queue import Queue
    q = Queue()
     
    # Enqueue root and initialize
    # height
    q.put(root)
     
    while(True):
 
        # nodeCount (queue size) indicates
        # number of nodes at current level
        nodeCount = q.qsize()
        if nodeCount == 0:
            break
 
        # Dequeue all nodes of current
        # level and Enqueue all nodes
        # of next level  
        while nodeCount > 0:
            node = q.get()
            print node.data,
            if node.left is not None:
                q.put(node.left)
            if node.right is not None:
                q.put(node.right)
            nodeCount -= 1
 
        print
         
# Driver code
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.right.left = Node(4)
root.right.right = Node(5)
 
print "Level order traversal of given tree"
printLevelOrder(root)
 
root = flipBinaryTree(root)
 
print "\nLevel order traversal of the flipped tree"
printLevelOrder(root)
 
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#




// C# program to flip a binary tree
using System;
using System.Collections.Generic;
 
public class FlipTree
{
 
    // method to flip the binary tree
    public static Node flipBinaryTree(Node root)
    {
        if (root == null)
            return root;
        if (root.left == null && root.right ==null)
            return root;
 
        // recursively call the same method
        Node flippedRoot = flipBinaryTree(root.left);
 
        // rearranging main root Node after returning
        // from recursive call
        root.left.left = root.right;
        root.left.right = root;
        root.left = root.right = null;
        return flippedRoot;
    }
 
    // Iterative method to do level order traversal
    // line by line
    public static void printLevelOrder(Node root)
    {
        // Base Case
        if(root == null)
            return ;
         
        // Create an empty queue for level order traversal
        Queue<Node> q = new Queue<Node>();
         
        // Enqueue Root and initialize height
        q.Enqueue(root);
        while(true)
        {
            // nodeCount (queue size) indicates number
            // of nodes at current lelvel.
            int nodeCount = q.Count;
            if (nodeCount == 0)
                break;
             
            // Dequeue all nodes of current level and
            // Enqueue all nodes of next level
            while (nodeCount > 0)
            {
                Node node = q.Dequeue();
                Console.Write(node.data+" ");
                if (node.left != null)
                    q.Enqueue(node.left);
                if (node.right != null)
                    q.Enqueue(node.right);
                nodeCount--;
            }
            Console.WriteLine();
        }
    }
 
    // Driver code
    public static void Main(String []args)
    {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(1);
        root.right.left = new Node(4);
        root.right.right = new Node(5);
        Console.WriteLine("Level order traversal of given tree");
        printLevelOrder(root);
     
        root = flipBinaryTree(root);
        Console.WriteLine("Level order traversal of flipped tree");
        printLevelOrder(root);
    }
}
 
/* A binary tree node structure */
public class Node
{
    public int data;
    public Node left, right;
    public Node(int data)
    {
        this.data = data;
    }
};
 
// This code is contributed by Princi Singh

Output: 
 

Level order traversal of given tree
1 
2 3 
4 5 

Level order traversal of the flipped tree
2 
3 1 
4 5 


Iterative Approach 
This approach is contributed by Pal13
The iterative solution follows the same approach as the recursive one, the only thing we need to pay attention to is to save the node information that will be overwritten. 
 

C++




//  C/C++ program to flip a binary tree
#include <bits/stdc++.h>
using namespace std;
 
// A binary tree node structure
struct Node
{
    int data;
    Node *left, *right;
};
 
// Utility function to create a new Binary
// Tree Node
struct Node* newNode(int data)
{
    struct Node *temp = new struct Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
// method to flip the binary tree
Node* flipBinaryTree(Node* root)
{
    // Initialization of pointers
    Node *curr = root;
    Node *next = NULL;
    Node *temp = NULL;
    Node *prev = NULL;
     
    // Iterate through all left nodes
    while(curr)
    {
        next = curr->left;
         
        // Swapping nodes now, need temp to keep the previous right child
         
        // Making prev's right as curr's left child
        curr->left = temp;        
         
        // Storing curr's right child
        temp = curr->right;        
         
        // Making prev as curr's right child
        curr->right = prev;        
         
        prev = curr;
        curr = next;
    }
    return prev;
}
 
// Iterative method to do level order traversal
// line by line
void printLevelOrder(Node *root)
{
    // Base Case
    if (root == NULL) return;
 
    // Create an empty queue for level order traversal
    queue<Node *> q;
 
    // Enqueue Root and initialize height
    q.push(root);
 
    while (1)
    {
        // nodeCount (queue size) indicates number
        // of nodes at current lelvel.
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
 
        // Dequeue all nodes of current level and
        // Enqueue all nodes of next level
        while (nodeCount > 0)
        {
            Node *node = q.front();
            cout << node->data << " ";
            q.pop();
             
            if (node->left != NULL)
                q.push(node->left);
             
            if (node->right != NULL)
                q.push(node->right);
            nodeCount--;
        }
        cout << endl;
    }
}
 
// Driver code
int main()
{
     
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->right->left = newNode(4);
    root->right->right = newNode(5);
 
    cout << "Level order traversal of given tree\n";
    printLevelOrder(root);
 
    root = flipBinaryTree(root);
 
    cout << "\nLevel order traversal of the flipped"
            " tree\n";
    printLevelOrder(root);
    return 0;
}
 
// This article is contributed by Pal13

Java




// Java program to flip a binary tree
import java.util.*;
class GFG
{
     
// A binary tree node
static class Node
{
    int data;
    Node left, right;
};
 
// Utility function to create
// a new Binary Tree Node
 
static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// method to flip the binary tree
static Node flipBinaryTree(Node root)
{
    // Initialization of pointers
    Node curr = root;
    Node next = null;
    Node temp = null;
    Node prev = null;
     
    // Iterate through all left nodes
    while(curr != null)
    {
        next = curr.left;
         
        // Swapping nodes now, need
        // temp to keep the previous
        // right child
         
        // Making prev's right
        // as curr's left child
        curr.left = temp;        
         
        // Storing curr's right child
        temp = curr.right;        
         
        // Making prev as curr's
        // right child
        curr.right = prev;        
         
        prev = curr;
        curr = next;
    }
    return prev;
}
 
// Iterative method to do
// level order traversal
// line by line
static void printLevelOrder(Node root)
{
    // Base Case
    if (root == null) return;
 
    // Create an empty queue for
    // level order traversal
    Queue<Node> q = new LinkedList<Node>();
 
    // Enqueue Root and
    // initialize height
    q.add(root);
 
    while (true)
    {
        // nodeCount (queue size)
        // indicates number of nodes
        // at current lelvel.
        int nodeCount = q.size();
        if (nodeCount == 0)
            break;
 
        // Dequeue all nodes of current
        // level and Enqueue all nodes
        // of next level
        while (nodeCount > 0)
        {
            Node node = q.peek();
            System.out.print(node.data + " ");
            q.remove();
             
            if (node.left != null)
                q.add(node.left);
             
            if (node.right != null)
                q.add(node.right);
            nodeCount--;
        }
        System.out.println();
    }
}
 
// Driver code
public static void main(String args[])
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.right.left = newNode(4);
    root.right.right = newNode(5);
 
    System.out.print("Level order traversal " +
                            "of given tree\n");
    printLevelOrder(root);
 
    root = flipBinaryTree(root);
 
    System.out.print("\nLevel order traversal " +
                        "of the flipped tree\n");
    printLevelOrder(root);
}
}
 
// This code is contributed
// by Arnab Kundu

Python3




# Python3 program to flip
# a binary tree
from collections import deque
 
# A binary tree node structure
class Node:
   
    def __init__(self, key):
       
        self.data = key
        self.left = None
        self.right = None
 
# method to flip the
# binary tree
def flipBinaryTree(root):
   
    # Initialization of
    # pointers
    curr = root
    next = None
    temp = None
    prev = None
 
    # Iterate through all
    # left nodes
    while(curr):
        next = curr.left
 
        # Swapping nodes now, need temp
        # to keep the previous right child
 
        # Making prev's right as curr's
        # left child
        curr.left = temp
 
        # Storing curr's right child
        temp = curr.right
 
        # Making prev as curr's right
        # child
        curr.right = prev
 
        prev = curr
        curr = next
    return prev
 
# Iterative method to do level
# order traversal line by line
def printLevelOrder(root):
   
    # Base Case
    if (root == None):
        return
 
    # Create an empty queue for
    # level order traversal
    q = deque()
 
    # Enqueue Root and initialize
    # height
    q.append(root)
 
    while (1):
        # nodeCount (queue size) indicates
        # number of nodes at current level.
        nodeCount = len(q)
        if (nodeCount == 0):
            break
 
        # Dequeue all nodes of current
        # level and Enqueue all nodes
        # of next level
        while (nodeCount > 0):
            node = q.popleft()
            print(node.data, end = " ")
 
            if (node.left != None):
                q.append(node.left)
 
            if (node.right != None):
                q.append(node.right)
            nodeCount -= 1
 
        print()
 
# Driver code
if __name__ == '__main__':
   
    root = Node(1)
    root.left = Node(2)
    root.right = Node(3)
    root.right.left = Node(4)
    root.right.right = Node(5)
 
    print("Level order traversal of given tree")
    printLevelOrder(root)
 
    root = flipBinaryTree(root)
 
    print("\nLevel order traversal of the flipped"
          " tree")
    printLevelOrder(root)
 
# This code is contributed by Mohit Kumar 29

C#




// C# program to flip a binary tree
using System;
using System.Collections.Generic;
 
class GFG
{
 
// A binary tree node
public class Node
{
    public int data;
    public Node left, right;
}
 
// Utility function to create
// a new Binary Tree Node
public static Node newNode(int data)
{
    Node temp = new Node();
    temp.data = data;
    temp.left = temp.right = null;
    return temp;
}
 
// method to flip the binary tree
public static Node flipBinaryTree(Node root)
{
    // Initialization of pointers
    Node curr = root;
    Node next = null;
    Node temp = null;
    Node prev = null;
 
    // Iterate through all left nodes
    while (curr != null)
    {
        next = curr.left;
 
        // Swapping nodes now, need
        // temp to keep the previous
        // right child
 
        // Making prev's right
        // as curr's left child
        curr.left = temp;
 
        // Storing curr's right child
        temp = curr.right;
 
        // Making prev as curr's
        // right child
        curr.right = prev;
 
        prev = curr;
        curr = next;
    }
    return prev;
}
 
// Iterative method to do level
// order traversal line by line
public static void printLevelOrder(Node root)
{
    // Base Case
    if (root == null)
    {
        return;
    }
 
    // Create an empty queue for
    // level order traversal
    LinkedList<Node> q = new LinkedList<Node>();
 
    // Enqueue Root and
    // initialize height
    q.AddLast(root);
 
    while (true)
    {
        // nodeCount (queue size)
        // indicates number of nodes
        // at current lelvel.
        int nodeCount = q.Count;
        if (nodeCount == 0)
        {
            break;
        }
 
        // Dequeue all nodes of current
        // level and Enqueue all nodes
        // of next level
        while (nodeCount > 0)
        {
            Node node = q.First.Value;
            Console.Write(node.data + " ");
            q.RemoveFirst();
 
            if (node.left != null)
            {
                q.AddLast(node.left);
            }
 
            if (node.right != null)
            {
                q.AddLast(node.right);
            }
            nodeCount--;
        }
        Console.WriteLine();
    }
}
 
// Driver code
public static void Main(string[] args)
{
    Node root = newNode(1);
    root.left = newNode(2);
    root.right = newNode(3);
    root.right.left = newNode(4);
    root.right.right = newNode(5);
 
    Console.Write("Level order traversal " +
                         "of given tree\n");
    printLevelOrder(root);
 
    root = flipBinaryTree(root);
 
    Console.Write("\nLevel order traversal " +
                     "of the flipped tree\n");
    printLevelOrder(root);
}
}
 
// This code is contributed by Shrikant13

Output: 
 

Level order traversal of given tree
1 
2 3 
4 5 

Level order traversal of the flipped tree
2 
3 1 
4 5 

Complexity Analysis: 
Time complexity: O(n) as in the worst case, depth of binary tree will be n. 
Auxiliary Space: O(1).

https://www.youtube.com/watch?v=S8GZFCbst2g 
 

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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