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

从深度优先遍历的角度来看，每次遇到一个包含子节点中间双链表节点，就递归的调用展开方法将其展开，并将展开的结果插入到当前节点的后面。这里须要注意双链表前节点先后指针的变动。步骤以下：

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

代码以下：

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

上面的思路一样能够经过循环的方式来解决。每遇到一个有子节点的双链表节点，就将其子节点的头和尾拼接到父节点的双链表上，使其看上去是一个新的双链表。再对双链表的下一个节点进行判断。基本步骤以下：

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

代码以下：

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：减小遍历次数

以前的两种思路，都会出现大量的重复遍历，重复遍历和叶子节点的深度成正相关，能够想方法将重复遍历的次数减小。其实，咱们能够看见，不管咱们什么时候将子节点展开，并拼接回父节点的双链表中，子节点展开的双链表的头结点是固定的，而且能够用父节点访问到。而尾节点必须经过重复遍从来查找并拼接。所以，若是每次都将展开后的尾节点返回，就能够无需重复遍历将展开的子节点拼接回父节点。代码以下：

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;
        }
    }