leetcode430. Flatten a Multilevel Doubly Linked List

題目要求

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：遞歸實現深度優先遍歷

從深度優先遍歷的角度來看，每次遇到一個包含子節點中間雙鏈表節點，就遞歸的調用展開方法將其展開，並將展開的結果插入到當前節點的後面。這裏須要注意雙鏈表前節點先後指針的變動。步驟以下：

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

代碼以下：

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：循環

上面的思路一樣能夠經過循環的方式來解決。每遇到一個有子節點的雙鏈表節點，就將其子節點的頭和尾拼接到父節點的雙鏈表上，使其看上去是一個新的雙鏈表。再對雙鏈表的下一個節點進行判斷。基本步驟以下：

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

代碼以下：

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：減小遍歷次數

以前的兩種思路，都會出現大量的重複遍歷，重複遍歷和葉子節點的深度成正相關，能夠想方法將重複遍歷的次數減小。其實，咱們能夠看見，不管咱們什麼時候將子節點展開，並拼接回父節點的雙鏈表中，子節點展開的雙鏈表的頭結點是固定的，而且能夠用父節點訪問到。而尾節點必須經過重複遍從來查找並拼接。所以，若是每次都將展開後的尾節點返回，就能夠無需重複遍歷將展開的子節點拼接回父節點。代碼以下：

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;
        }
    }