Flatten a Multilevel Doubly Linked List

Description

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(javascript)

/*
 * @lc app=leetcode id=430 lang=javascript
 *
 * [430] Flatten a Multilevel Doubly Linked List
 */

// @lc code=start
/**
 * @param {Node} head
 * @return {Node}
 */
const flatten = function (head) {
  if (!head) {
    return null
  }
  const result = []
  const aux = (node) => {
    if (!node) {
      return
    }
    result.push(node)
    aux(node.child)
    aux(node.next)
  }
  aux(head)
  result.forEach((node, index) => {
    node.child = null
    if (result[index + 1]) {
      node.next = result[index + 1]
      result[index + 1].prev = node
      result[index + 1].next = null
    }
  })
  return result[0]
}