430. Flatten a Multilevel Doubly Linked List
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:
- he number of Nodes will not exceed 1000.
- 1 < = N o d e . v a l < = 1 0 5 1 <= Node.val <= 10^5 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]
From: LeetCode
Link: 430. Flatten a Multilevel Doubly Linked List
Solution:
Ideas:
1. Recursive Flattening: The function flatten is called recursively on any node that has a child list. The recursion ensures that we handle all nested child lists first, flattening them level by level.
2. Linking the Child List: Once the child list is flattened, we update the pointers to insert the child list between the current node (curr) and the next node (nextNode).
3. Handling the Next Node: After linking the child list, the tail of the child list is connected to the original next node, maintaining the structure of the doubly linked list.
4. Complexity Consideration:
- Time Complexity: Each node is visited once, so the time complexity is O(N), where N is the total number of nodes in the list.
- Space Complexity: The space complexity is O(N) due to the recursive stack in the worst case where all nodes are connected via child pointers.
Code:
/*
// Definition for a Node.
class Node {
public:
int val;
Node* prev;
Node* next;
Node* child;
};
*/
class Solution {
public:
Node* flatten(Node* head) {
if (!head) return head;
Node* curr = head;
while (curr) {
// If the current node has a child, we need to flatten it
if (curr->child) {
Node* nextNode = curr->next;
Node* childHead = flatten(curr->child); // Recursively flatten the child list
// Attach the child list to the current node
curr->next = childHead;
childHead->prev = curr;
// Find the tail of the child list
Node* tail = childHead;
while (tail->next) {
tail = tail->next;
}
// Attach the next part of the original list to the tail
tail->next = nextNode;
if (nextNode) {
nextNode->prev = tail;
}
// Nullify the child pointer after flattening
curr->child = nullptr;
}
// Move to the next node
curr = curr->next;
}
return head;
}
};