430. Flatten a Multilevel Doubly Linked List

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:

-------------------------
Solution #1, recursive

/*
// 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 null;
        Node cp = head;
        rec(cp);
        
        return head;
    }
    
    private Node rec(Node head) {
        if (head.next == null && head.child == null) return head;
        
        if (head.child == null) {
            return rec(head.next);
        }
        
        Node savedNext = head.next;
        head.next = head.child;
        head.child.prev = head;

        Node last = rec(head.child);
        head.child = null;
        last.next = savedNext;
        if (savedNext != null) {
            savedNext.prev = last;
        }
        savedNext = last;
        return rec(savedNext);
    }
}

Solution #2, iterative. 如果有child，就把child整个list并回主list(暂时先放过child的child)，然后再从child开始走。每次都减少一层
O(n), 因为每个node最多被走了2次

/*
// 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) {
        Node itr = head;
        
        while (itr != null) {
            if (itr.child != null) {
                Node savedNext = itr.next;
                itr.next = itr.child;
                itr.child.prev = itr;
                itr.child = null;
                
                Node itrInner = itr.next;
                while (itrInner.next != null) {
                    itrInner = itrInner.next;
                }
                
                itrInner.next = savedNext;
                if (savedNext != null) savedNext.prev = itrInner;
            }
            
            itr = itr.next;
        }
        
        return head;
    }
}