430. Flatten a Multilevel Doubly Linked List

https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/

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

把三向链表（额外带个child）压缩成双向的，对有child的把next后面的接到child尾部，并用child替换next。除了直接实现题意的方法，因为child下还有child，属于嵌套定义，可以用stack存下遇到带child结点的next到栈，然后把child替换到next，继续遍历。当next为空时从stack里拿并接上。

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

// @lc code=start
/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* prev;
    Node* next;
    Node* child;

    Node() {}

    Node(int _val, Node* _prev, Node* _next, Node* _child) {
        val = _val;
        prev = _prev;
        next = _next;
        child = _child;
    }
};
*/
class Solution {
public:
    Node* flatten(Node* head) {
        stack<Node*> s;
        Node *cur = head;
        while (cur != nullptr) {
            if (cur->child != nullptr) {
                if (cur->next != nullptr) s.push(cur->next);
                cur->next = cur->child;
                cur->child->prev = cur;
                cur->child = nullptr;
            }
            if (cur->next == nullptr && !s.empty()) {
                cur->next = s.top();
                cur->next->prev = cur;
                s.pop();
            }  
            cur = cur->next;
        }
        return head;
    }
};
// @lc code=end