430 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:

 1 <--> 2 <--> 3 <--> 4 <--> 5 <--> 6
               |
               7 <--> 8 <--> 9 <--> 10
                      |
                      11 <--> 12
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:
 1 <--> 2 <--> 3 <--> 7 <--> 8 <--> 11 <--> 12 <--> 9 <--> 10 <--> 4 <--> 5 <--> 6

Example 2:

 1 <--> 2
 |
 3
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:
1 <--> 3 <--> 2

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]

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"""
# Definition for a Node.
class Node:
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child
"""

class Solution:
    def flatten(self, head: 'Optional[Node]') -> 'Optional[Node]':
        if not head:
            return head
        stack = []
        cur = head
        while cur or stack:
            if cur.child:
                if cur.next:
                    stack.append(cur.next)
                cur.next = cur.child
                cur.child.prev = cur
                cur.child = None
            elif not cur.next and stack:
                cur.next = stack.pop()
                cur.next.prev = cur
            cur = cur.next
        return head

"""
DFS
"""
class Solution(object):

    def flatten(self, head: 'Optional[Node]') -> 'Optional[Node]':
        if not head:
            return head

        pseudoHead = Node(None, None, head, None)
        self.flatten_dfs(pseudoHead, head)
        
        pseudoHead.next.prev = None
        return pseudoHead.next


    def flatten_dfs(
        self,
        prev: 'Optional[Node]',
        curr: 'Optional[Node]'
    ) -> 'Optional[Node]':
        """ return the tail of the flatten list """
        if not curr:
            return prev

        curr.prev = prev
        prev.next = curr

        # the curr.next would be tempered in the recursive function
        tempNext = curr.next
        tail = self.flatten_dfs(curr, curr.child)
        curr.child = None
        return self.flatten_dfs(tail, tempNext)