Remove Linked List Elements
<p>Given the <code>head</code> of a linked list and an integer <code>val</code>, remove all the nodes of the linked list that has <code>Node.val == val</code>, and return <em>the new head</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0203.Remove%20Linked%20List%20Elements/images/removelinked-list.jpg" style="width: 500px; height: 142px;" />
<pre>
<strong>Input:</strong> head = [1,2,6,3,4,5,6], val = 6
<strong>Output:</strong> [1,2,3,4,5]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> head = [], val = 1
<strong>Output:</strong> []
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> head = [7,7,7,7], val = 7
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 50</code></li>
	<li><code>0 &lt;= val &lt;= 50</code></li>
</ul>
Recursion; Linked List
class Solution {
public:
    ListNode* removeElements(ListNode* head, int val) {
        ListNode* dummy = new ListNode();
        dummy->next = head;
        ListNode* p = dummy;
        while (p->next) {
            if (p->next->val == val) {
                p->next = p->next->next;
            } else {
                p = p->next;
            }
        }
        return dummy->next;
    }
};