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:

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// 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;
}

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



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