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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 singlelevel, 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 <= 10^{5}
How the multilevel linked list is represented in test cases:
We use the multilevel linked list from Example 1 above:
123456NULL

78910NULL

1112NULL
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]
class Solution {
public Node flatten(Node head) {
if( head == null) return head;
// Pointer
Node p = head;
while( p!= null) {
/* CASE 1: if no child, proceed */
if( p.child == null ) {
p = p.next;
continue;
}
/* CASE 2: got child, find the tail of the child and link it to p.next */
Node temp = p.child;
// Find the tail of the child
while( temp.next != null )
temp = temp.next;
// Connect tail with p.next, if it is not null
temp.next = p.next;
if( p.next != null ) p.next.prev = temp;
// Connect p with p.child, and remove p.child
p.next = p.child;
p.child.prev = p;
p.child = null;
}
return head;
}
}
Node* flatten(Node* head) {
for (Node* h = head; h; h = h>next)
{
if (h>child)
{
Node* next = h>next;
h>next = h>child;
h>next>prev = h;
h>child = NULL;
Node* p = h>next;
while (p>next) p = p>next;
p>next = next;
if (next) next>prev = p;
}
}
return head;
}
class Solution(object):
def flatten(self, head):
if not head:
return
dummy = Node(0,None,head,None)
stack = []
stack.append(head)
prev = dummy
while stack:
root = stack.pop()
root.prev = prev
prev.next = root
if root.next:
stack.append(root.next)
root.next = None
if root.child:
stack.append(root.child)
root.child = None
prev = root
dummy.next.prev = None
return dummy.next
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