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;
    public Node prev;
    public Node next;
    public Node child;
};
*/

class Solution {
      
public Node flatten(Node head) {
        helper(head);
        return head;
    }

     Node helper(Node head) {
        if (head == null) return head; // CASE 1
        if (head.child == null) {
            if (head.next == null) return head; // CASE 2
            return helper(head.next); // CASE 3
        }
        else {
            Node child = head.child;  
            head.child = null;
            Node next = head.next;
            Node childtail = helper(child);
            head.next = child;
            child.prev = head;  
            if (next != null) { // CASE 5
                childtail.next = next;
                next.prev = childtail;
                return helper(next);
            }
            return childtail; // CASE 4
        }           
    }
}

Try it on Leetcode

Here, we are going to solve using normal recursion. Once child node found we need to change next of origin node and prev of child node. Tail represents last node. Helper method will flatten head and return the tail in its child(if any exists)

Below are cases and these are done in above code.,

CASE 1. null - no need to flatten, just return it
CASE 2. no child, no next - no need to flatten, it is the last element, just return it
CASE 3. no child, next - no need to flatten, go next
CASE 4. child, no next - flatten the child and done
CASE 5. child, next - flatten the child, connect it with the next, go next

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