Flatten a Multilevel Doubly Linked List

Flatten a Multilevel Doubly Linked List

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

 /*

// Definition for a Node.

class Node {

public:

    int val;

    Node* prev;

    Node* next;

    Node* child;

};

*/

class Solution {

public:

//     Node* flat(Node* h)

//     {

//         Node* curr=h,*tail=h;

//         while(curr)

//         {

//             Node *child=curr->child;

//             Node *next=curr->next;

//             if(child)

//             {

//                 tail=flat(child);

                

//                 tail->next=next;

//                 if(next)

//                 next->prev=tail;

                

                

//                 curr->next=child;

//                 child->prev=curr;

                

//                 curr->child=NULL;

//                 curr=tail;

                

//             }

//             curr=next;

//                 if(curr)

//                     tail=curr;

//         }

        

//         return tail;

//     }

    Node* flatten(Node* head) {

        // if(head)flat(head);

        // return head;

        if (!head) return NULL;

        Node *current = head, *next;

        while(current) {

            if (current->child) {

                next = current->next;

                Node* child = flatten(current->child);

                current->child = NULL;

                current->next = child;

                child->prev = current;

                Node *p = child;

                while(p && p->next)

                    p = p->next;

                p->next = next;

                if (next)

                    next->prev = p;

            }

            current = current->next;

        }

        

        return head;

        

    }

};