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 singlelevel, doubly linked list. You are given the head of the first level of the list.
Example:
Input:
123456NULL

78910NULL

1112NULL
Output:
123781112910456NULL
Explanation for the above example:
Given the following multilevel doubly linked list:
We should return the following flattened doubly linked list:
对于任意一个节点，处理的顺序如下，
child
子串，并将其嫁接在本节点之后next
子串，并将其嫁接在child
子串之后/*
// Definition for a Node.
class Node {
public int val;
public Node prev;
public Node next;
public Node child;
public Node() {}
public Node(int _val,Node _prev,Node _next,Node _child) {
val = _val;
prev = _prev;
next = _next;
child = _child;
}
};
*/
class Solution {
public Node flatten(Node head) {
if (head == null) return null;
Node next = flatten(head.next);
Node child = flatten(head.child);
head.child = null;
head.next = child;
if (child != null) child.prev = head;
Node oldHead = head;
while (head != null && head.next != null) head = head.next;
head.next = next;
if (next != null) next.prev = head;
return oldHead;
}
}