« back

Flatten a Multilevel Doubly Linked List

Tags : linkedlist, leetcode, cpp, medium

Problem Statement - link #

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.

Examples #

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 #

Solutions #

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* prev;
    Node* next;
    Node* child;
};
*/

class Solution {
public:
    Node* flatten(Node* head) {
      Node* tmp = head;
      vector<Node*> nodes;
      int level = 1;
      if(!head) return head;
      
      while(1){
        
        if(tmp->child != NULL)
          nodes.push_back(tmp);
          
        if(tmp->next == NULL)
          if(level <= nodes.size()){
            tmp = nodes[nodes.size()-1]->child;
            level++;
          }
          else
            break;
        
        else
          tmp = tmp->next;
        
        cout<<tmp->val<<endl;
      }
      for(int i = nodes.size()-1; i>=0; i--){
        Node* r = nodes[i]->next;
        Node* c = nodes[i]->child;
        nodes[i]->child = NULL;
        nodes[i]->next = c;
        c->prev = nodes[i];
        while(c->next != NULL) c = c->next;
        c->next = r;
        (r!=NULL ? r->prev = c : r);
      }
      return head;
    }
};