Flatten a Multilevel Doubly Linked List
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
Note
迭代法:
直接改吧,找到有child的节点curr,找到这一层的tail,连到curr的后面(注意断开或者新连接都是双向的)对于不同层次的在下次会被访问,相当于level一直在减少
递归法:
对于链表,递归执行顺序不一样,返回第一层的时候,下面几层都已经搞好了。递归的参数往往是下一个节点。帮助函数返回必须得是tail了这里,所以需要一个帮助函数,因为主函数返回的是head。
Code
public Node flatten(Node head) {
if (head == null) {
return head;
}
Node curr = head;
while (curr != null) {
if (curr.child == null) {
curr = curr.next;
continue;
} else {
Node c = curr.child;
Node next = curr.next;
curr.next = c;
c.prev = curr;
while (c.next != null) {
c = c.next;
}
if (next != null) {
c.next = next;
next.prev = c;
}
curr.child = null;
}
}
return head;
}
public Node flatten(Node head) {
helper(head);
return head;
}
private Node helper(Node head) {
Node cur = head, pre = head;
while (cur != null) {
if (cur.child == null) {
pre = cur;
cur = cur.next;
} else {
Node tmp = cur.next;
Node child = helper(cur.child);
cur.next = cur.child;
cur.child.prev = cur;
cur.child = null;
child.next = tmp;
if (tmp != null) tmp.prev = child;
pre = child;
cur = tmp;
}
}
return pre;
}
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