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:
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]
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[ 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]
In this approach, we utilize a stack to achieve depth-first traversal of the multilevel doubly linked list. We push nodes into the stack starting from the head, along with managing the child nodes as higher priority over next nodes. This ensures that we process all child nodes before moving on to the next nodes.
The C solution employs a stack data structure to manage the traversal of the multilevel doubly linked list. We start at the head and check if there's a child. If there is, we push the next pointer, set the child as the next node, and nullify the child. If we encounter the end of a list and have stored nodes in the stack, we pop one to continue.
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Time Complexity: O(n) where n is the number of nodes. Each node is visited once.
Space Complexity: O(n) for the stack in the worst case scenario.
This approach utilizes recursion to handle the traversing and flattening of lists. By inherently using the function call stack, it efficiently manages shifts between the parent and child lists, automatically flattening the entire structure as it recursively resolves each node and its children.
The recursive C solution uses a helper function inside the primary function to handle recursive flattening, where each call processes and flattens the child list before linking back to the parent node’s next list.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n) due to the necessity to visit each node once.
Space Complexity: O(d) where d is the maximum depth of the children, necessitating stack space for recursion.
| Approach | Complexity |
|---|---|
| Iterative Approach Using Stack | Time Complexity: O(n) where n is the number of nodes. Each node is visited once. |
| Recursive Approach | Time Complexity: O(n) due to the necessity to visit each node once. |
Flatten Binary Tree to Linked List - Leetcode 114 - Python • NeetCode • 56,714 views views
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