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# Flatten a Multilevel Doubly Linked List LeetCode Solution

## Problem – Flatten a Multilevel Doubly Linked List

You are given a doubly linked list, which contains nodes that have a next pointer, a previous pointer, and an additional child pointer. This child pointer may or may not point to a separate doubly linked list, also containing these special nodes. 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.

Given the `head` of the first level of the list, flatten the list so that all the nodes appear in a single-level, doubly linked list. Let `curr` be a node with a child list. The nodes in the child list should appear after `curr` and before `curr.next` in the flattened list.

Return the `head` of the flattened list. The nodes in the list must have all of their child pointers set to `null`.

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 shown.
After flattening the multilevel linked list it becomes:
``````

Example 2:

``````Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation: The multilevel linked list in the input is shown.
After flattening the multilevel linked list it becomes:
``````

Example 3:

``````Input: head = []
Output: []
Explanation: There could be empty list in the input.
``````

Constraints:

• The number of Nodes will not exceed `1000`.
• `1 <= Node.val <= 105`

How the multilevel linked list is represented in test cases:

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]
``````

### Flatten a Multilevel Doubly Linked List LeetCode Solution in Java

``````class Solution {
// Pointer
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;
}
}
}
``````

### Flatten a Multilevel Doubly Linked List LeetCode Solution in C++

``````Node* flatten(Node* head) {
for (Node* h = head; h; h = h->next)
{
if (h->child)
{
Node* next = h->next;
h->next = h->child;
h->next->prev = h;
h->child = NULL;
Node* p = h->next;
while (p->next) p = p->next;
p->next = next;
if (next) next->prev = p;
}
}
}
``````

### Flatten a Multilevel Doubly Linked List LeetCode Solution in Python

``````class Solution(object):
return

stack = []
prev = dummy

while stack:
root = stack.pop()

root.prev = prev
prev.next = root

if root.next:
stack.append(root.next)
root.next = None
if root.child:
stack.append(root.child)
root.child = None
prev = root

dummy.next.prev = None
return dummy.next
``````
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