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:
- The number of Nodes will not exceed
1000. 1 <= Node.val <= 105
"""
# 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);
}
}