Given the head
of a linked list and an integer val
, remove all the nodes of the linked list that has Node.val == val
, and return the new head.
Example 1:
Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]
Example 2:
Input: head = [], val = 1
Output: []
Example 3:
Input: head = [7,7,7,7], val = 7
Output: []
Constraints:
- The number of nodes in the list is in the range
[0, 104]
. -
1 <= Node.val <= 50
-
0 <= val <= 50
SOLUTION:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
if not head:
return head
if head.val == val:
return self.removeElements(head.next, val)
else:
head.next = self.removeElements(head.next, val)
return head
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