flatten-a-multilevel-doubly-linked-list.py

'''

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:

Input:
 1---2---3---4---5---6--NULL
         |
         7---8---9---10--NULL
             |
             11--12--NULL

Output:
1-2-3-7-8-11-12-9-10-4-5-6-NULL


Explanation for the above example:

Given the following multilevel doubly linked list:




We should return the following flattened doubly linked list:




'''



"""
# 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: 'Node') -> 'Node':

        # Approach one
        # key = []        # 运用栈结构
        # res = head
        # while head:
        #     if head.child:
        #         key.append(head.next)
        #         head.next = head.child
        #         head.next.prev = head
        #         head.child = None
        #     if not head.next and key:
        #         key_note = key.pop()
        #         while not key_note and key:    # 防止子链出现在父链的末尾结点
        #             key_note = key.pop()
        #         if not key_note: break
        #         head.next = key_note
        #         head.next.prev = head
        #     head = head.next
        # return res



        # Approach two    简化上述代码，不会入栈空结点
        stack = []
        res = head
        while head:
            if head.child:
                if head.next:
                    stack.append(head.next)
                head.next, head.child.prev, head.child = head.child, head, None
            elif head.next == None and stack:
                node = stack.pop()
                head.next, node.prev = node, head
            head = head.next
        return res