leetcode430

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。这里需要注意双链表前节点前后指针的变更。步骤如下：

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
        

代码如下：

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

上面的思路同样可以通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。基本步骤如下：

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
        

代码如下：

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：减少遍历次数

之前的两种思路，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，可以想方法将重复遍历的次数减少。其实，我们可以看见，无论我们何时将子节点展开，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，并且可以用父节点访问到。而尾节点必须通过重复遍历来查找并拼接。因此，如果每次都将展开后的尾节点返回，就可以无需重复遍历将展开的子节点拼接回父节点。代码如下：

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;
    }
}