'''
430. Flatten a Multilevel Doubly Linked List        Medium

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:
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child
"""

class Solution:
    '''
    This solution is pretty straightforward. We need to go through this linked list. 
    Whenever we encounter a node that has a child, we save this node's next (if exists) to a list.
    Then we set current node's next to be its child, the child's previous to be current node. 
    If current node has no next, we check the list. If it is empty, we are done. Otherwise, 
    we set the last added node as current's node's next and continue. 
    
    Complexity: time O(n); space O(n) where n is the total number of nodes in the linked list. 
    '''
    def flatten(self, head: 'Node') -> 'Node':
        d = []
        node = head
        while node:
            if node.child:
                if node.next: d.append(node.next)
                node.next = node.child
                node.child.prev = node
                node.child = None
                node = node.next
            else: 
                if not node.next:
                    if not d: break
                    popped = d.pop()
                    node.next = popped
                    popped.prev = node
                node = node.next
        return head
                
e8e629c22c4a9dc0484128cb17caa80fbd7ad1ec
<mask token>

<mask token>


class Solution:
    <mask token>
    <mask token>

<mask token>


class Solution:
    <mask token>

    def flatten(self, head: 'Node') ->'Node':
        d = []
        node = head
        while node:
            if node.child:
                if node.next:
                    d.append(node.next)
                node.next = node.child
                node.child.prev = node
                node.child = None
                node = node.next
            else:
                if not node.next:
                    if not d:
                        break
                    popped = d.pop()
                    node.next = popped
                    popped.prev = node
                node = node.next
        return head

<mask token>


class Solution:
    """
    This solution is pretty straightforward. We need to go through this linked list. 
    Whenever we encounter a node that has a child, we save this node's next (if exists) to a list.
    Then we set current node's next to be its child, the child's previous to be current node. 
    If current node has no next, we check the list. If it is empty, we are done. Otherwise, 
    we set the last added node as current's node's next and continue. 
    
    Complexity: time O(n); space O(n) where n is the total number of nodes in the linked list. 
    """

    def flatten(self, head: 'Node') ->'Node':
        d = []
        node = head
        while node:
            if node.child:
                if node.next:
                    d.append(node.next)
                node.next = node.child
                node.child.prev = node
                node.child = None
                node = node.next
            else:
                if not node.next:
                    if not d:
                        break
                    popped = d.pop()
                    node.next = popped
                    popped.prev = node
                node = node.next
        return head

'''
430. Flatten a Multilevel Doubly Linked List        Medium

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:
    def __init__(self, val, prev, next, child):
        self.val = val
        self.prev = prev
        self.next = next
        self.child = child
"""

class Solution:
    '''
    This solution is pretty straightforward. We need to go through this linked list. 
    Whenever we encounter a node that has a child, we save this node's next (if exists) to a list.
    Then we set current node's next to be its child, the child's previous to be current node. 
    If current node has no next, we check the list. If it is empty, we are done. Otherwise, 
    we set the last added node as current's node's next and continue. 
    
    Complexity: time O(n); space O(n) where n is the total number of nodes in the linked list. 
    '''
    def flatten(self, head: 'Node') -> 'Node':
        d = []
        node = head
        while node:
            if node.child:
                if node.next: d.append(node.next)
                node.next = node.child
                node.child.prev = node
                node.child = None
                node = node.next
            else: 
                if not node.next:
                    if not d: break
                    popped = d.pop()
                    node.next = popped
                    popped.prev = node
                node = node.next
        return head
                