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.
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:
Input: head = [1,2,null,3] Output: [1,3,2] Explanation: The input multilevel linked list is as follows: 1---2---NULL | 3---NULL
Input: head = [] Output: []
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] - Number of Nodes will not exceed 1000.
1 <= Node.val <= 10^5
# Definition for a Node.# class Node# attr_accessor :val, :prev, :next, :child# def initialize(val=nil, prev=nil, next_=nil, child=nil)# @val = val# @prev = prev# @next = next_# @child = child# end# end# @param {Node} root# @return {Node}defflatten(root)returnnilifroot.nil?prev=Node.newstack=[root]untilstack.empty?curr=stack.popstack.push(curr.next)unlesscurr.next.nil?stack.push(curr.child)unlesscurr.child.nil?prev.next=currcurr.prev=prevcurr.child=nilprev=currendroot.prev=nilreturnrootend