day-101.cpp

/*
  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* flattenHelper(Node* head) {
        Node* curr = head;
        Node* tail = head;

        while (curr) {
            Node* child = curr->child;
            Node* next = curr->next;

            if (child) {
                Node* _tail = flattenHelper(child);
                _tail->next = next;
                if (next) next->prev = _tail;

                curr->next = child;
                child->prev = curr;

                curr->child = nullptr;

                curr = tail;
            } else {
                curr = next;
            }

            if (curr) tail = curr;
        }

        return tail;
    }

    Node* flatten(Node* head) {
        if (head) flattenHelper(head);
        return head;
    }
};