leetcode_gym/_problems/flattenMultilevelDoublyLinkedList.md

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