LeetCode #430. Flatten a Multilevel Doubly Linked List

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本文介绍了一种算法，用于将多级双向链表展平为单级双向链表。通过递归方法，该算法能有效处理链表中可能存在的多级子链表结构，确保所有节点位于同一级别。

题目描述：

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:

We should return the following flattened doubly linked list:

class Solution {
public:
    Node* flatten(Node* head) {
        if(head==NULL) return NULL;
        Node* p=head;
        // 从头结点开始，找到一个有子节点的节点
        while(p!=NULL&&p->child==NULL) p=p->next;
        // 如果找不到，直接返回
        if(p==NULL) return head;
        
        // 如果找到了，先把子节点压平，放置于当前节点之后
        Node* q=p->next;
        Node* tmp=flatten(p->child);
        p->next=tmp;
        p->child=NULL;
        tmp->prev=p;
        
        // 剩余的节点压平再接在后面
        while(p!=NULL&&p->next!=NULL) p=p->next;
        tmp=flatten(q);
        p->next=tmp;
        if(tmp) tmp->prev=p;
        return head;
    }
};