LeetCode 430 Flatten A Multilevel Doubly Linked List - Medium

430. Flatten A Multilevel Doubly Linked List — Medium

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Problem

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

Solution

### using stack:
# 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':
        if not head:
            return head
        cur = head
        stack = []
        while cur:
            if cur.child:
                if cur.next:
                    stack.append(cur.next)
                cur.next = cur.child
                cur.child.prev = cur
                cur.child = None
            else:
                if cur.next is None and stack:
                    cur.next = stack.pop()
                    cur.next.prev = cur
            cur = cur.next
        return head