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
func removeElements(head *ListNode, val int) *ListNode {
	dummy := new(ListNode)
	dummy.Next = head
	p := dummy
	for p.Next != nil {
		if p.Next.Val == val {
			p.Next = p.Next.Next
		} else {
			p = p.Next
		}
	}
	return dummy.Next
}