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* flatten(Node* head) {

        if (head == NULL) {

            return head;

        }

        Node *trav = head;

        while (trav != NULL) {

            if (trav -> child == NULL) {

                trav = trav -> next;

                continue;

            }

            Node *flat = NULL, *flat_last = NULL;

            flat = flatten(trav -> child);

            Node *trav_next = trav -> next;

            trav -> next = flat;

            flat -> prev = trav;

            

            // Otherwise go forward to find the flat_last.

            if (flat_last == NULL) {

                Node *trav2 = flat;

                while (trav2 -> next != NULL) {

                    trav2 = trav2 -> next;

                }

                flat_last = trav2;

            }

            // Attach flattened last to trav_next.

            if (trav_next != NULL) {

                // Handle last.

                flat_last -> next = trav_next;

                trav_next -> prev = flat_last;

            } else {

                flat_last -> next = NULL;

            }

            trav -> child = NULL;

            trav = trav_next;

        }

        return head;

    }

};