# LeetCode – Flatten a Multilevel Doubly Linked List

## 题目：

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:

```Input:
1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL

Output:
1-2-3-7-8-11-12-9-10-4-5-6-NULL```

## 解题：

1. 先判断child节点是否为空，如果是存在；否则继续遍历；
2. 如果存在child，将当前节点的next节点保存， 将当前节点的next指向child节点， 设置child的prev为当前节点，清空child节点；
2. 找到child链表的末尾节点；
3. child末尾节点指向保存的next节点， next节点指向child节点的末尾节点。

## 实现：

``````/*
// 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:
while (cur!=NULL){
if (cur->child!=NULL){
Node* next = cur->next;
cur->next = cur->child;
cur->child->prev = cur;
cur->child = NULL;
Node *p = cur->next;
while(p->next!=NULL&&p!=NULL) {
p= p->next;
}
p->next = next;
if (next!=NULL){
next->prev = p;
}
}
cur= cur->next;
}