You are given a doubly linked list, which contains nodes that have a next pointer, a previous pointer, and an additional child pointer. This child pointer may or may not point to a separate doubly linked list, also containing these special nodes. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure as shown in the example below.

Given the head of the first level of the list, flatten the list so that all the nodes appear in a single-level, doubly linked list. Let curr be a node with a child list. The nodes in the child list should appear after curr and before curr.next in the flattened list.

Return the head of the flattened list. The nodes in the list must have all of their child pointers set to null.

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Example 1:

Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
Output: [1,2,3,7,8,11,12,9,10,4,5,6]
Explanation: The multilevel linked list in the input is shown.
After flattening the multilevel linked list it becomes:



Example 2:

Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation: The multilevel linked list in the input is shown.
After flattening the multilevel linked list it becomes:



Example 3:

Input: head = []
Output: []
Explanation: There could be empty list in the input.


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Constraints:


	The number of Nodes will not exceed 1000.
	1 &lt;= Node.val &lt;= 105


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How the multilevel linked list is represented in test cases:

We use the multilevel linked list from Example 1 above:

 1---2---3---4---5---6--NULL
         |
         7---8---9---10--NULL
             |
             11--12--NULL

The serialization of each level is as follows:

[1,2,3,4,5,6,null]
[7,8,9,10,null]
[11,12,null]


To serialize all levels together, we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:

[1,    2,    3, 4, 5, 6, null]
             |
[null, null, 7,    8, 9, 10, null]
                   |
[            null, 11, 12, null]


Merging the serialization of each level and removing trailing nulls we obtain:

[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]


"""
# Definition for a Node.
class Node:
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child
"""

class Solution:
    def flatten(self, head: 'Optional[Node]') -> 'Optional[Node]':    
        node = head
        while node:
            if node.child: # If there is a child travel to last node of the child
                child = node.child
                while child.next:
                    child = child.next
                child.next = node.next # Update the next of child to the the next of the current node
                if node.next: # update the prev of the next node to chile to make it valid doubly linked list
                    node.next.prev = child
                node.next = node.child # Update the child to become the next of the current
                node.next.prev = node # update the prev of the next node to chile to make it valid doubly linked list
                node.child = None # Make the child of the current node None to fulfill the requirements
            node = node.next
        return head

# time and space complexity
# time: O(n)
# space: O(1)
class Solution {
    public Node flatten(Node head) {
        Node curr = head ; // for traversal
        Node tail = head; // for keeping the track of previous node
        Stack<Node> stack = new Stack<>(); // for storing the reference of next node when child node encounters
        while(curr != null){
            if(curr.child != null){ // if there is a child
                Node child = curr.child; // creating a node for child
                if(curr.next != null){ // if there is list after we find child a child
                    stack.push(curr.next); // pushing the list to the stack
                    curr.next.prev = null; // pointing its previous to null
                }
                curr.next = child; // pointing the current's reference to child
                child.prev = curr; // pointing child's previous reference to current.
                curr.child = null; // pointing the current's child pointer to null
            }
            tail = curr ; // for keeping track of previous nodes
            curr= curr.next; // traversing
        }
        while(!stack.isEmpty()){ // checking if the stack has still nodes in it.
            curr = stack.pop(); // getting the last node of the list pushed into the stack
            tail.next = curr; // pointing the previos node to the last node
            curr.prev = tail; // pointing previos pointer of the last node to the previos node.
            while( curr != null){ // traversing the last node's popped out of stack
                tail = curr;
                curr = curr.next ;
            }
        }
        return head;
    }
}
class Solution {
public:
    Node* flatten(Node* head)
    {
        if(head==NULL) return head;
       Node *temp=head;
       stack<Node*> stk;
       while(temp->next!=NULL || temp->child!=NULL  || stk.size()!=0)
       {
           if(temp->next==NULL && temp->child==NULL && stk.size())
           {
               Node *a=stk.top();
               stk.pop();
               temp->next=a;
               a->prev=temp;
           }
           if(temp->child!=NULL)
           {
               if(temp->next!=NULL)
               {
               Node* a=temp->next;
               a->prev=NULL;
               stk.push(a);
               }
               temp->next=temp->child;
               temp->next->prev=temp;
               temp->child=NULL;
               
           }
           temp=temp->next;
       }
        return head;
    }
};
Feel free to ask in doubt in comment section
var flatten = function(head) {
    var arr = [];
    var temp = head;
    var prev= null;
    while(temp)
        {
            if(temp.child!= null)
                {
                    arr.push(temp.next);
                    temp.next = temp.child;
                    temp.child.prev = temp;
                    temp.child = null;
                }
            prev = temp;
            temp = temp.next
        }
    for(var j=arr.length-1; j>=0; j--)
        {
            if(arr[j] != null)
                mergeOtherLists(arr[j]);
        }
    return head;
	
	function mergeOtherLists(root)
		{
			prev.next=root;
			root.prev=prev;
		   while(root)
			   {
				  prev = root;
				   root = root.next;
			   }
		}
};
Flatten a Multilevel Doubly Linked List