430. Flatten a Multilevel Doubly Linked List--Medium

ゝ一纸荒年。 2023-08-17 15:58 104阅读 0赞

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

Explanation for the above example:

Given the following multilevel doubly linked list:  
![1632149-20190904171012229-1749964750.png][]

We should return the following flattened doubly linked list:  
![1632149-20190904171022910-2123879355.png][]

1.思考

*  该方法主要有以下几个重点：  
    (1). 首先想到利用递归的方法深度搜索；  
    (2). 其次保留本节点原先的下一跳节点rightList，即同一层右边的list；  
    (3). 再将本节点与child建立双向连接；  
    (4). 然后child的返回节点得是最后一个元素节点，而非NULL，所以用preNode保存curNode的上一跳节点；  
    (5). 将child返回的节点与righList建立双向连接，这里要考虑rightList是否是一层的最后结尾。

2.实现  
Runtime: 88ms(95%)  
Memory: 31MB(100%)

/*
    // Definition for a Node.
    class Node {
    public:
        int val;
        Node* prev;
        Node* next;
        Node* child;
    
        Node() {}
    
        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 *pHead = head;
            Node *curNode = FlattenList(head);
            return pHead;
        }
        
        Node* FlattenList(Node* head){
            Node *curNode = head, *preNode=NULL;
            while(curNode!=NULL){
                if(curNode->child!=NULL){
                    //先保留同一层右边的list
                    Node *rightList = curNode->next;
                    //rightList->prev = NULL;
                    //再与child节点建立双向指针
                    curNode->next = curNode->child;
                    curNode->next->prev = curNode;
                    curNode->child = NULL;
                    //深度搜索本节点的child
                    curNode = FlattenList(curNode->next);
                    //将child的最后一个节点与本节点的下一跳相连
                    if(rightList==NULL){
                        return curNode;
                    }
                    curNode->next = rightList;
                    rightList->prev = curNode;
                }
                //preNode为curNode的上一跳节点，防止curNode=NULL
                preNode = curNode;
                curNode = curNode->next;
            }
            return preNode;
        }
    };

转载于:https://www.cnblogs.com/xuyy-isee/p/11460358.html

[1632149-20190904171012229-1749964750.png]: /images/20230810/2664cfb2c14e4a648eca3c35c0e36e07.png
[1632149-20190904171022910-2123879355.png]: /images/20230810/cf07ad4ef20749228c5354d010c57f30.png