Remove linked list elements

Give you a list of the head node head And an integer val , Please delete all the contents in the linked list Node.val == val The node of , And back to New head node .

One of the ideas ：
Be careful ：
Consider that the linked list is empty , The running result is null , There is only one node in the linked list
The wrong sample ：

#include<stdio.h>
#include<stdlib.h>
struct ListNode 
{
    
    int val;
    struct ListNode *next; 
};
struct ListNode* removeElements(struct ListNode* head, int val){
    
	struct ListNode* cur, *newhead, *tail;
	tail = NULL;
	newhead = NULL;
	cur = head;
	if (head == NULL)
		return NULL;
	while (cur)
	{
    
		if (cur->val != val)
		{
    
			if (newhead == NULL)
			{
    
				newhead = cur;
				tail = cur;
				cur = cur->next;
			}
			else
			{
    
				tail->next = cur;
				tail = cur;
				cur = cur->next;
			}
		}
		else
		{
    
			cur = cur->next;
		}
	}
	if (tail)
		tail->next = NULL;
	return newhead;
}
int main()
{
    
	struct ListNode * n1 = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode * n2 = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode * n3 = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode * n4 = (struct ListNode*)malloc(sizeof(struct ListNode));
	n1->val = 1;
	n2->val = 2;
	n3->val = 3;
	n4->val = 4;
	n1->next = n2;
	n2->next = n3;
	n3->next = n4;
	n4->next = NULL;

	struct ListNode* plist = n1;
	removeElements(plist,3);

}

The above code can be through oj, But the problem of memory leakage needs to be solved , Code style can also be optimized .

The correct sample

#include<stdio.h>
#include<stdlib.h>
struct ListNode
{
    
	int val;
	struct ListNode *next;
};
struct ListNode* removeElements(struct ListNode* head, int val){
    
	if (head == NULL)
		return NULL;
	struct ListNode* cur = head;
	struct ListNode* newhead = NULL, *tail = NULL;
	while (cur)
	{
    
		// If the value is val, need free Corresponding node ,
		// Corresponding next Also released , It is difficult to find the next node , So save the next node 
		struct ListNode* next = cur->next;
		if (cur->val != val)
		{
    
			if (newhead == NULL)
			{
    
				newhead = cur;
				tail = cur;
			}
			else
			{
    
				tail->next = cur;
				tail = cur;
			}
		}
		else
		{
    
			free(cur);
		}
		cur = next;
	}
	if (tail)
		tail->next = NULL;
	return newhead;
}
int main()
{
    
	struct ListNode * n1 = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode * n2 = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode * n3 = (struct ListNode*)malloc(sizeof(struct ListNode));
	struct ListNode * n4 = (struct ListNode*)malloc(sizeof(struct ListNode));
	n1->val = 1;
	n2->val = 2;
	n3->val = 3;
	n4->val = 4;
	n1->next = n2;
	n2->next = n3;
	n3->next = n4;
	n4->next = NULL;

	struct ListNode* plist = n1;
	removeElements(plist, 3);

}

Running results
Pictured
Post operation node n3 Be released , Linked list becomes n1->n2->n4->NULL