LeetCode刷题实战430：扁平化多级双向链表-技术圈

算法的重要性，我就不多说了吧，想去大厂，就必须要经过基础知识和业务逻辑面试+算法面试。所以，为了提高大家的算法能力，这个公众号后续每天带大家做一道算法题，题目就从LeetCode上面选！

今天和大家聊的问题叫做 扁平化多级双向链表，我们先来看题面：

https://leetcode-cn.com/problems/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.

多级双向链表中，除了指向下一个节点和前一个节点指针之外，它还有一个子链表指针，可能指向单独的双向链表。这些子列表也可能会有一个或多个自己的子项，依此类推，生成多级数据结构，如下面的示例所示。

给你位于列表第一级的头节点，请你扁平化列表，使所有结点出现在单级双链表中。

示例

解题

思路：采用DFS，注释都写在代码里了。

class Solution {
public:
    Node* flatten(Node* head) {
        Node *cur = head, *nextNode = NULL, *prevN = head;
        //所有的指针记得初始化，不要以为后面要付值就不要初始值，怕后面if没进去
        while(prevN)
        { //遍历链表
            if(cur && cur->child) //当前有，且有子链
            {
                nextNode = cur->next;//记住上层cur后面的
                cur->next = flatten(cur->child);//展平
                cur->child->prev = cur;//子链头接上，展平
                cur->child = NULL;//将cur孩子删掉
            }
            prevN = cur;//prevnode移动
            if(cur)
                cur = cur->next;//cur移动
            if(!cur && nextNode)
            { //当cur到终点的下一个时，且上层后面还有nextnode
                prevN->next = nextNode;//展平的终点prevN指向上层的nextnode
                nextNode->prev = prevN;//反向接上
                cur = prevN->next;//cur移动
                nextNode = NULL;//已接好，暂时没有child的需要展平，nextNode清空
            }
        }
        return head;
    }
};