<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://assets.leetcode.com/uploads/2021/03/06/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>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next

def node_gen(start, val):
    if start.next:
        ptr=start.next
        while ptr:
            if ptr.val != val:
                yield ptr
            ptr=ptr.next
    else:
        ...

class Solution:
    def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]:
        start=head
        while start and (start.val == val and start.next):
            start=start.next
        if start is None or start.val==val:
            return
        head_ptr=ListNode(start.val,start.next)
        ptr=head_ptr
        print(list(node.val for node in node_gen(start,val)))
        for node in node_gen(start,val):
            ptr.next=node
            ptr=node
        ptr.next=None
        return head_ptr
                
        
18900