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]
Your algorithm should be iterative (non-recursive) and run in O(n) time.
The core challenge of this problem is to flatten a multilevel doubly linked list into a single-level doubly linked list. This involves traversing through each node and ensuring that any child nodes are integrated into the main list in the correct order.
Common applications of this problem include data structures that need to be linearized for easier traversal or processing, such as hierarchical file systems or nested organizational charts.
Potential pitfalls include not correctly handling the child pointers or losing track of the next pointers during the flattening process.
To solve this problem, we can use an iterative approach with a stack to keep track of nodes. This ensures that we can handle the multilevel nature of the list without using recursion, which could lead to stack overflow for very deep lists.
1. **Naive Solution**: A naive solution might involve recursively traversing the list, but this is not optimal due to potential stack overflow issues.
2. **Optimized Solution**: An iterative solution using a stack is more efficient and avoids the pitfalls of recursion. This approach ensures that we can handle any depth of nesting without running into stack overflow issues.
The algorithm can be broken down into the following steps:
class Node:
def __init__(self, val, prev=None, next=None, child=None):
self.val = val
self.prev = prev
self.next = next
self.child = child
def flatten(head):
if not head:
return head
# Initialize the stack and push the head of the list onto the stack
stack = []
stack.append(head)
prev = None
while stack:
curr = stack.pop()
# Link the current node to the previous node
if prev:
prev.next = curr
curr.prev = prev
# If the current node has a next pointer, push it onto the stack
if curr.next:
stack.append(curr.next)
# If the current node has a child pointer, push it onto the stack
if curr.child:
stack.append(curr.child)
curr.child = None
# Move the previous pointer to the current node
prev = curr
return head
The time complexity of this algorithm is O(n), where n is the number of nodes in the list. This is because each node is processed exactly once.
The space complexity is O(n) in the worst case, where all nodes are pushed onto the stack.
Potential edge cases include:
Each of these cases is handled by the algorithm as it processes each node individually and correctly handles the child pointers.
To test the solution comprehensively, we can use a variety of test cases, from simple to complex:
def print_list(head):
result = []
while head:
result.append(head.val)
head = head.next
return result
# Test case 1
head = Node(1, None, Node(2, None, Node(3, None, Node(4, None, Node(5, None, Node(6))))))
head.next.next.child = Node(7, None, Node(8, None, Node(9, None, Node(10))))
head.next.next.child.next.child = Node(11, None, Node(12))
flattened = flatten(head)
print(print_list(flattened)) # Output: [1, 2, 3, 7, 8, 11, 12, 9, 10, 4, 5, 6]
# Test case 2
head = Node(1, None, Node(2))
head.child = Node(3)
flattened = flatten(head)
print(print_list(flattened)) # Output: [1, 3, 2]
# Test case 3
head = None
flattened = flatten(head)
print(print_list(flattened)) # Output: []
When approaching such problems, it's important to:
Practicing similar problems and studying different algorithms can help improve problem-solving skills.
Flattening a multilevel doubly linked list is a common problem that can be efficiently solved using an iterative approach with a stack. Understanding the problem, breaking it down into smaller steps, and considering edge cases are key to developing an effective solution.
By practicing and exploring further, you can improve your problem-solving skills and become more proficient in handling complex data structures.
For further reading and practice problems, consider the following resources: