# Flatten Binary Tree in order of Level Order Traversal

• Difficulty Level : Basic
• Last Updated : 17 Dec, 2021

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 ``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 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 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 q = ``new` `Queue();``    ``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

 ``

Output :

`1 5 2 6 4 9 3 `

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

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