Question

430. Flatten a Multilevel Doubly Linked List

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Description

You are given a doubly linked list, which contains nodes that have a next pointer, a previous pointer, and an additional child pointer. This child pointer may or may not point to a separate doubly linked list, also containing these special nodes. 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.

Given the head of the first level of the list, flatten the list so that all the nodes appear in a single-level, doubly linked list. Let curr be a node with a child list. The nodes in the child list should appear after curr and before curr.next in the flattened list.

Return _the _head_ of the flattened list. The nodes in the list must have all of their child pointers set to _null.

 

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 shown.
After flattening the multilevel linked list it becomes:

Example 2:

Input: head = [1,2,null,3]
Output: [1,3,2]
Explanation: The multilevel linked list in the input is shown.
After flattening the multilevel linked list it becomes:

Example 3:

Input: head = []
Output: []
Explanation: There could be empty list in the input.

 

Constraints:

• The number of Nodes will not exceed 1000.
• 1 <= Node.val <= 105

 

How the multilevel linked list is represented in test cases:

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]

Solutions

Solution 1

"""
# Definition for a Node.
class Node:
  def __init__(self, val, prev, next, child):
    self.val = val
    self.prev = prev
    self.next = next
    self.child = child
"""


class Solution:
  def flatten(self, head: 'Node') -> 'Node':
    def preorder(pre, cur):
      if cur is None:
        return pre
      cur.prev = pre
      pre.next = cur

      t = cur.next
      tail = preorder(cur, cur.child)
      cur.child = None
      return preorder(tail, t)

    if head is None:
      return None
    dummy = Node(0, None, head, None)
    preorder(dummy, head)
    dummy.next.prev = None
    return dummy.next

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

class Solution {
  public Node flatten(Node head) {
    if (head == null) {
      return null;
    }
    Node dummy = new Node();
    dummy.next = head;
    preorder(dummy, head);
    dummy.next.prev = null;
    return dummy.next;
  }

  private Node preorder(Node pre, Node cur) {
    if (cur == null) {
      return pre;
    }
    cur.prev = pre;
    pre.next = cur;

    Node t = cur.next;
    Node tail = preorder(cur, cur.child);
    cur.child = null;
    return preorder(tail, t);
  }
}

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

class Solution {
public:
  Node* flatten(Node* head) {
    flattenGetTail(head);
    return head;
  }

  Node* flattenGetTail(Node* head) {
    Node* cur = head;
    Node* tail = nullptr;

    while (cur) {
      Node* next = cur->next;
      if (cur->child) {
        Node* child = cur->child;
        Node* childTail = flattenGetTail(cur->child);

        cur->child = nullptr;
        cur->next = child;
        child->prev = cur;
        childTail->next = next;

        if (next)
          next->prev = childTail;

        tail = childTail;
      } else {
        tail = cur;
      }

      cur = next;
    }

    return tail;
  }
};