430. 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.

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:

M1: straight forward

用一个pointer遍历linked list，遇到有child的节点，就把整个child linked list加入linked list，再继续遍历，直到走到null为止

time: O(n)　　-- n: total # of nodes, space: O(1)

/*
// Definition for a Node.
class Node {
    public int val;
    public Node prev;
    public Node next;
    public Node child;

    public Node() {}

    public Node(int _val,Node _prev,Node _next,Node _child) {
        val = _val;
        prev = _prev;
        next = _next;
        child = _child;
    }
};
*/
class Solution {
    public Node flatten(Node head) {
        if(head == null) {
            return head;
        }
        Node cur = head;
        while(cur != null) {
            if(cur.child == null) {
                cur = cur.next;
                continue;
            }
            Node childTail = cur.child;
            while(childTail.next != null) {
                childTail = childTail.next;
            }
            
            if(cur.next != null) {
                cur.next.prev = childTail;
            }
            childTail.next = cur.next;
            
            cur.next = cur.child;
            cur.child.prev = cur;
            cur.child = null;
            cur = cur.next;
        }
        return head;
    }
}

M2: recursion, dfs

1. head = null，或者只有head一个节点，没next/child，return；

2. 没child, next；

3. 有child，没next，flatten即可；

4. 有child，有next，flatten完，把cur.next和child的tail连接起来

time: O(n)　　-- n: total # of nodes, space: O(k)　　-- k: total # of child

/*
// Definition for a Node.
class Node {
    public int val;
    public Node prev;
    public Node next;
    public Node child;

    public Node() {}

    public Node(int _val,Node _prev,Node _next,Node _child) {
        val = _val;
        prev = _prev;
        next = _next;
        child = _child;
    }
};
*/
class Solution {
    public Node flatten(Node head) {
        dfs(head);
        return head;
    }
    
    private Node dfs(Node head) {
        if(head == null) {
            return head;
        } else if(head.child == null) {
            if(head.next == null) {
                return head;
            }
            return dfs(head.next);
        } else {
            Node child = head.child;
            head.child = null;
            Node next = head.next;
        
            Node childTail = dfs(child);
            head.next = child;
            child.prev = head;
        
            if(next != null) {
                childTail.next = next;
                next.prev = childTail;
                return dfs(next);
            }
        
            return childTail;
        }
    }
}