Flatten Binary Tree in order of Level Order Traversal

Given a Binary Tree, the task is to flatten it in order of Level order traversal of the tree. In the flattened binary tree, the left node of all the nodes must be NULL.
Examples:

Input: 
          1 
        /   \ 
       5     2 
      / \   / \ 
     6   4 9   3 
Output: 1 5 2 6 4 9 3

Input:
      1
       \
        2
         \
          3
           \
            4
             \
              5
Output: 1 2 3 4 5

Approach: We will solve this problem by simulating the Level order traversal of Binary Tree as follows: 

1. Create a queue to store the nodes of Binary tree.
2. Create a variable “prev” and initialise it by parent node.
3. Push left and right children of parent in the queue.
4. Apply level order traversal. Lets say “curr” is front most element in queue. Then, 
   • If ‘curr’ is NULL, continue.
   • Else push curr->left and curr->right in the queue
   • Set prev = curr

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Node of the Binary tree
struct node {
    int data;
    node* left;
    node* right;
    node(int data)
    {
        this->data = data;
        left = NULL;
        right = NULL;
    }
};

// Function to flatten Binary tree using
// level order traversal
void flatten(node* parent)
{
    // Queue to store nodes
    // for BFS
    queue<node*> q;
    q.push(parent->left);
    q.push(parent->right);
    node* prev = parent;

    // Code for BFS
    while (q.size()) {

        // Size of queue
        int s = q.size();
        while (s--) {

            // Front most node in
            // the queue
            node* curr = q.front();
            q.pop();

            // Base case
            if (curr == NULL)
                continue;
            prev->right = curr;
            prev->left = NULL;
            prev = curr;

            // Pushing new elements
            // in queue
            q.push(curr->left);
            q.push(curr->right);
        }
    }

    prev->left = NULL;
    prev->right = NULL;
}

// Function to print flattened
// Binary Tree
void print(node* parent)
{
    node* curr = parent;
    while (curr != NULL)
        cout << curr->data << " ", curr = curr->right;
}

// Driver code
int main()
{
    node* root = new node(1);
    root->left = new node(5);
    root->right = new node(2);
    root->left->left = new node(6);
    root->left->right = new node(4);
    root->right->left = new node(9);
    root->right->right = new node(3);

    // Calling required functions
    flatten(root);
    print(root);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Node of the Binary tree
static class node
{
    int data;
    node left;
    node right;
    node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
};

// Function to flatten Binary tree using
// level order traversal
static void flatten(node parent)
{
    // Queue to store nodes
    // for BFS
    Queue<node> q = new LinkedList<>();
    q.add(parent.left);
    q.add(parent.right);
    node prev = parent;

    // Code for BFS
    while (q.size() > 0)
    {

        // Size of queue
        int s = q.size();
        while (s-- > 0)
        {

            // Front most node in
            // the queue
            node curr = q.peek();
            q.remove();

            // Base case
            if (curr == null)
                continue;
            prev.right = curr;
            prev.left = null;
            prev = curr;

            // Pushing new elements
            // in queue
            q.add(curr.left);
            q.add(curr.right);
        }
    }
    prev.left = null;
    prev.right = null;
}

// Function to print flattened
// Binary Tree
static void print(node parent)
{
    node curr = parent;
    while (curr != null)
    {
        System.out.print(curr.data + " ");
        curr = curr.right;
    }
}

// Driver code
public static void main(String[] args)
{
    node root = new node(1);
    root.left = new node(5);
    root.right = new node(2);
    root.left.left = new node(6);
    root.left.right = new node(4);
    root.right.left = new node(9);
    root.right.right = new node(3);

    // Calling required functions
    flatten(root);
    print(root);
}
}

// This code is contributed by Rajput-Ji

Python3

# Python implementation of above algorithm

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

# Function to flatten Binary tree using
# level order traversal
def flatten( parent):

    # Queue to store nodes
    # for BFS
    q = []
    q.append(parent.left)
    q.append(parent.right)
    prev = parent

    # Code for BFS
    while (len(q) > 0) :

        # Size of queue
        s = len(q)
        while (s > 0) :
            s = s - 1

            # Front most node in
            # the queue
            curr = q[0]
            q.pop(0)

            # Base case
            if (curr == None):
                continue
            prev.right = curr
            prev.left = None
            prev = curr

            # appending elements
            # in queue
            q.append(curr.left)
            q.append(curr.right)

    prev.left = None
    prev.right = None

# Function to print flattened
# Binary Tree
def print_(parent):

    curr = parent
    while (curr != None):
        print( curr.data , end=" ")
        curr = curr.right

# Driver code
root = node(1)
root.left = node(5)
root.right = node(2)
root.left.left = node(6)
root.left.right = node(4)
root.right.left = node(9)
root.right.right = node(3)

# Calling required functions
flatten(root)
print_(root)

# This code is contributed by Arnab Kundu

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;

class GFG
{

// Node of the Binary tree
public class node
{
    public int data;
    public node left;
    public node right;
    public node(int data)
    {
        this.data = data;
        left = null;
        right = null;
    }
};

// Function to flatten Binary tree using
// level order traversal
static void flatten(node parent)
{
    // Queue to store nodes
    // for BFS
    Queue<node> q = new Queue<node>();
    q.Enqueue(parent.left);
    q.Enqueue(parent.right);
    node prev = parent;

    // Code for BFS
    while (q.Count > 0)
    {

        // Size of queue
        int s = q.Count;
        while (s-- > 0)
        {

            // Front most node in
            // the queue
            node curr = q.Peek();
            q.Dequeue();

            // Base case
            if (curr == null)
                continue;
            prev.right = curr;
            prev.left = null;
            prev = curr;

            // Pushing new elements
            // in queue
            q.Enqueue(curr.left);
            q.Enqueue(curr.right);
        }
    }
    prev.left = null;
    prev.right = null;
}

// Function to print flattened
// Binary Tree
static void print(node parent)
{
    node curr = parent;
    while (curr != null)
    {
        Console.Write(curr.data + " ");
        curr = curr.right;
    }
}

// Driver code
public static void Main(String[] args)
{
    node root = new node(1);
    root.left = new node(5);
    root.right = new node(2);
    root.left.left = new node(6);
    root.left.right = new node(4);
    root.right.left = new node(9);
    root.right.right = new node(3);

    // Calling required functions
    flatten(root);
    print(root);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// Javascript implementation of the approach

// Node of the Binary tree
class node
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}

// Function to flatten Binary tree using
// level order traversal
function flatten(parent)
{
    // Queue to store nodes
    // for BFS
    let q = [];
    q.push(parent.left);
    q.push(parent.right);
    let prev = parent;

    // Code for BFS
    while (q.length > 0)
    {

        // Size of queue
        let s = q.length;
        while (s-- > 0)
        {

            // Front most node in
            // the queue
            let curr = q.shift();


            // Base case
            if (curr == null)
                continue;
            prev.right = curr;
            prev.left = null;
            prev = curr;

            // Pushing new elements
            // in queue
            q.push(curr.left);
            q.push(curr.right);
        }
    }
    prev.left = null;
    prev.right = null;
}

// Function to print flattened
// Binary Tree
function print(parent)
{
    let curr = parent;
    while (curr != null)
    {
        document.write(curr.data + " ");
        curr = curr.right;
    }
}

// Driver code
let root = new node(1);
root.left = new node(5);
root.right = new node(2);
root.left.left = new node(6);
root.left.right = new node(4);
root.right.left = new node(9);
root.right.right = new node(3);

// Calling required functions
flatten(root);
print(root);

// This code is contributed by avanitrachhadiya2155
</script>

Output :

1 5 2 6 4 9 3 

Time Complexity: O(N) 
Space Complexity: O(N) where N is the size of Binary Tree.