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]

Constraints:

Number of Nodes will not exceed 1000. 1 <= Node.val <= 10^5

Pre analysis

Will simply iterate and find child recursively.

Another Solution

Rather simple iteration

var flatten = function(head) { let curr = head while(curr){ if(curr.child) { let child = curr.child let currNext = curr.next

          curr.next = child
          child.prev = curr
          
          while(child.next) child = child.next
          
          child.next = currNext
          if(currNext) currNext.prev = child        
      }
      
      curr.child = null        
      curr = curr.next
  }
  
  return head   };