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]
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
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36/*
// 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) {
Node p = head;
while (p != null) {
if (p.child != null) {
Node right = p.next;
// process child
p.next = flatten(p.child);
p.next.prev = p;
p.child = null;
while (p.next != null) {
p = p.next;
}
// reconnect next
if (right != null) {
p.next = right;
p.next.prev = p;
}
}
p = p.next;
}
return head;
}
}