430. Flatten a Multilevel Doubly Linked List ---M

 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

A:

就是一个个递归，然后插入即可。

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/* // Definition for a Node. class Node { public: int val; Node* prev; Node* next; Node* child; }; */ class Solution { public: Node* flatten(Node* head) { Node* runner = head; while(runner!=nullptr){ if(runner->child){ Node* subHead = flatten(runner->child); runner->child = nullptr; // insert subHead after runner Node* nextNode = runner->next; runner->next = subHead; subHead->prev = runner; while(runner->next){ // till the end runner=runner->next; } runner->next = nextNode; if(nextNode) nextNode->prev = runner; } runner = runner->next; } return head; } };

Mistakes：

28行， 没有检查 nextNode