---
layout: page
title:  Flatten a Multilevel Doubly Linked List
last_solved: 2020-07-10
category: linked list
leetcode_url: https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/
status: Attempted
---

Problem
-------

```
You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. 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.

Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.

 

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 as follows:



After flattening the multilevel linked list it becomes:


Example 2:

Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation:

The input multilevel linked list is as follows:

  1---2---NULL
  |
  3---NULL
Example 3:

Input: head = []
Output: []
 

How multilevel linked list is represented in test case:

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]

```

Solution
----------

{% highlight python %}

def flatten(self, head: 'Node') -> 'Node':
    
    
    if not head: return head
    
    dummy = Node(0)
    
    curr = dummy
    stack = [head]
    
    while stack:
        
        tmp = stack.pop()
        
        if tmp.next: stack.append( tmp.next )
        if tmp.child: stack.append( tmp.child )
        
        print( tmp.val )
        
        curr.next = tmp     
        tmp.prev = curr
        
        tmp.child = None    # no child pointers
        curr = tmp          # move ahead
    
    dummy.next.prev = None
    return dummy.next

{% endhighlight %}


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