430. Flatten a Multilevel Doubly Linked List

1. Question

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.

2. Examples

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]

3. Constraints

• The number of Nodes will not exceed 1000.
• 1 <= Node.val <= 10⁵

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/flatten-a-multilevel-doubly-linked-list 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

4. Solutions

主要的地方还是在dfs的时候检测到子节点，需要对下一个节点进行压栈，在子节点完了后，需要弹栈。对拼接的地方next和prev进行赋值。

class Solution {
  public Node flatten(Node head) {
    if (head == null) {
      return null;
    }

    Stack<Node> stack = new Stack<>();
    Node ans = head;

    while (head != null) {
      if (head.child != null) {
        if (head.next != null) {
          stack.push(head.next);
        }
        head.next = head.child;
        head.next.prev = head;
        head.child = null;
      } else if (head.next == null && !stack.isEmpty()) {
        head.next = stack.pop();
        head.next.prev = head;
      }
      head = head.next;
    }
    return ans;
  }
}