430.Flatten-a-Multilevel-Doubly-Linked-List

430. Flatten a Multilevel Doubly Linked List

题目地址

https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/

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 1:
Input: head = [1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]
Output: [1,2,3,7,8,11,12,9,10,4,5,6]
Explanation:

The multilevel linked list in the input is as follows:

After flattening the multilevel linked list it becomes:

Example 2:
Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation:

The input multilevel linked list is as follows:

  1---2---NULL
  |
  3---NULL

Example 3:
Input: head = []
Output: []

How multilevel linked list is represented in test case:

We use the multilevel linked list from Example 1 above:

 1---2---3---4---5---6--NULL
         |
         7---8---9---10--NULL
             |
             11--12--NULL
The serialization of each level is as follows:

[1,2,3,4,5,6,null]
[7,8,9,10,null]
[11,12,null]
To serialize all levels together we will add nulls in each level to signify no node connects to the upper node of the previous level. The serialization becomes:

[1,2,3,4,5,6,null]
[null,null,7,8,9,10,null]
[null,11,12,null]
Merging the serialization of each level and removing trailing nulls we obtain:

[1,2,3,4,5,6,null,null,null,7,8,9,10,null,null,11,12]

Constraints:
Number of Nodes will not exceed 1000.
1 <= Node.val <= 10^5

代码

Approach #1 DFS by Recursion

/*
// 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;
    // pseudo head to ensure the `prev` pointer is never none
    Node pseudoHead = new Node(0, null, head, null);

    flattenDFS(pseudoHead, head);

    // detach the pseudo head from the real head
    pseudoHead.next.prev = null;
    return pseudoHead.next;
  }
  /* return the tail of the flatten list */
  public Node flattenDFS(Node prev, Node curr) {
    if (curr == null) return prev;
    curr.prev = prev;
    prev.next = curr;

    // the curr.next would be tempered in the recursive function
    Node tempNext = curr.next;

    Node tail = flattenDFS(curr, curr.child);
    curr.child = null;

    return flattenDFS(tail, tempNext);
  }
}

Approach #2 DFS by Iteration

Complexity

• Time Complexity: O(N). The iterative solution has the same time complexity as the recursive.

• Space Complexity: O(N). Again, the iterative solution has the same space complexity as the recursive one.

/*
// 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 pseudoHead = new Node(0, null, head, null);
    Deque<Node> stack = new ArrayDeque<>();
    stack.push(head);

    Node curr, prev = pseudoHead;
    while (!stack.isEmpty()) {
      curr = stak.pop();
      prev.next = curr;
      curr.prev = prev;

      if (curr.next != null)        stack.push(curr.next);
      if (curr.child != null) {
        stack.push(curr.child);
        curr.child = null;
      }
      prev = curr;
    }

    pseudoHead.next.prev = null;
    return pseudoHead.next;
  }
}