Day-10-Flatten a Multilevel Doubly Linked List.cpp

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) {
        Node *curr=head, *tail=head;
        stack<Node*> s;
        while(curr!=NULL){
            if(curr->child!=NULL){
                if(curr->next!=NULL){
                    s.push(curr->next);
                    curr->next->prev=NULL;
                }
                Node* child=curr->child;
                curr->next=child;
                child->prev=curr;
                curr->child=NULL;
            }
            tail=curr;
            curr=curr->next;
        }
        while(!s.empty()){
            curr=s.top();
            s.pop();
            tail->next=curr;
            curr->prev=tail;
            while(curr!=NULL){
                tail=curr;
                curr=curr->next;
            }
        }
        return head;
    }
};