text: /*
  430. Flatten a Multilevel Doubly Linked List
  Medium
  
  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:
      The number of Nodes will not exceed 1000.
      1 <= Node.val <= 105
  */
  /*
  // Definition for a Node.
  class Node {
  public:
      int val;
      Node* prev;
      Node* next;
      Node* child;
  };
  */
  
  class Solution {
  public:
      Node* flatten(Node* head) {
          Node* h=head,*next,*p;
          while(h){
              if(h->child){
                  next=h->next;
                  h->next=h->child;
                  h->next->prev=h;
                  h->child=NULL;
                  p=h->next;
                  while(p->next) p=p->next;
                  p->next=next;
                  if(next)next->prev=p;
              }
              h=h->next;
          }
          return head;
      }
  };
  