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

class Solution {
    public Node flatten(Node head) {
        if (head == null)
            return head;
        // Pointer
        Node p = head;
        while (p != null) {
            /* CASE 1: if no child, proceed */
            if (p.child == null) {
                p = p.next;
                continue;
            }
            /* CASE 2: got child, find the tail of the child and link it to p.next */
            Node temp = p.child;
            // Find the tail of the child
            while (temp.next != null)
                temp = temp.next;
            // Connect tail with p.next, if it is not null
            temp.next = p.next;
            if (p.next != null)
                p.next.prev = temp;
            // Connect p with p.child, and remove p.child
            p.next = p.child;
            p.child.prev = p;
            p.child = null;
        }
        return head;
    }
}