# leetcode430. Flatten a Multilevel Doubly Linked List

## 题目要求

``````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``````

## 思路一：递归实现深度优先遍历

``````Step1:
1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL

Step2:
1---2---3---4---5---6--NULL
|
7---8---11--12--9---10--NULL

Step3:
1---2---3---7---8---11--12--9---10--4---5---6--NULL
``````

``````    public Node flatten(Node head) {
if(head == null) return head;
Node tmp = head;
while(tmp != null) {
if(tmp.child != null) {
Node child = flatten(tmp.child);
tmp.child = null;
Node next = tmp.next;
tmp.next = child;
child.prev = tmp;
while(child.next != null) {
child = child.next;
}
child.next = next;
if(next != null) {
next.prev = child;
}
tmp = next;
}else {
tmp = tmp.next;

}
}
return head;
}``````

## 思路二：循环

``````Step1:
1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL

Step2:
1---2---3---7---8---9---10---4---5---6--NULL
|
11--12--NULL

Step3:
1---2---3---7---8---11--12--9---10--4---5---6--NULL
``````

``````    public Node flatten(Node head) {
if(head == null) return null;

Node tmp = head;
while(tmp != null) {
if(tmp.child != null) {

Node child = tmp.child;
tmp.child = null;

Node next = tmp.next;
tmp.next = child;
child.prev = tmp;
while(child.next != null) {
child =  child.next;
}

if(next != null) {
child.next = next;
next.prev = child;
}
}
tmp = tmp.next;
}
return head;
}``````

## 思路3：减少遍历次数

``````    public Node flatten(Node head) {
flattenAndReturnTail(head);
return head;
}

public Node flattenAndReturnTail(Node head) {
if(head == null) return null;
if(head.child == null) {
if(head.next == null) return head;
return flattenAndReturnTail(head.next);
}else {
Node child = head.child;
head.child = null;

Node next = head.next;
Node childTail = flatten(child);
head.next  = child;
child.prev = head;
if(next != null) {
childTail.next = next;
next.prev = childTail;
return flattenAndReturnTail(next);
}
return childTail;
}
}``````