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

time complexity : O(n)
space complexity : O(1)

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* prev;
    Node* next;
    Node* child;
};
*/

class Solution {
public:
    Node* flatten(Node* head) {
        if (!head) return nullptr;
        head->prev = nullptr;
        
        stack<Node*> preLv;
        for (Node* i = head; i!=nullptr; i=i->next) {
            if (i->child) {
                // go to next level
                preLv.push(i->next);
                i->child->prev = i;
                i->next = i->child;
                i->child = nullptr;
            } else if (i->next==nullptr && !preLv.empty()) {
                // reach end, back to prev level.
                i->next = preLv.top();
                preLv.pop();
                if (i->next) i->next->prev = i;
            }
        }
        
        return head;
    }
};

用一個指標從頭跑到尾，用一個stack存放該返回的node。
思考「往下一層」與「返回上一層」的處理。