05 August 2008
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
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class Solution {
Node result = new Node();
Node tail = result;
List<Integer> list = new ArrayList<>();
public Node flatten(Node head) {
if(head == null) {return head;}
Node cur = head;
flattenHead(cur);
tail.next = null;
result.next.prev = null;
return result.next;
}
private void flattenHead(Node node){
List<Integer> n = list;
if(node != null){
list.add(node.val);
tail.next = new Node(node.val);
tail.next.prev = tail;
tail = tail.next;
if(node.child != null){
flattenHead(node.child);
flattenHead(node.next);
} else{
flattenHead(node.next);
}
}
}
}