Description
Difficulty: Medium
Tag: LinkedList, DFS
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
Solution
从上到下,遇到child先flat child,然后将所有的连起来。
/*
// 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) {
Node first = new Node();
doFlatten(first, head);
Node result = first.next;
if(result != null){
result.prev = null;
}
return result;
}
private Node doFlatten(Node flatten, Node head) {
if (head == null) {
return flatten;
}
flatten.next = head;
head.prev = flatten;
Node tail = flatten.next;
Node next = head.next;
if (head.child != null) {
tail = doFlatten(tail, head.child);
head.child=null;
}
return doFlatten(tail, next);
}
}
来源:oschina
链接:https://my.oschina.net/u/1246890/blog/2874485