<p>Given the <code>head</code> of a linked list, we repeatedly delete consecutive sequences of nodes that sum to <code>0</code> until there are no such sequences.</p>

<p>After doing so, return the head of the final linked list.&nbsp; You may return any such answer.</p>

<p>&nbsp;</p>
<p>(Note that in the examples below, all sequences are serializations of <code>ListNode</code> objects.)</p>

<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> head = [1,2,-3,3,1]
<strong>Output:</strong> [3,1]
<strong>Note:</strong> The answer [1,2,1] would also be accepted.
</pre>

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

<pre>
<strong>Input:</strong> head = [1,2,3,-3,4]
<strong>Output:</strong> [1,2,4]
</pre>

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

<pre>
<strong>Input:</strong> head = [1,2,3,-3,-2]
<strong>Output:</strong> [1]
</pre>

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

<ul>
	<li>The given linked list will contain between <code>1</code> and <code>1000</code> nodes.</li>
	<li>Each node in the linked list has <code>-1000 &lt;= node.val &lt;= 1000</code>.</li>
</ul>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def removeZeroSumSublists(self, head: Optional[ListNode]) -> Optional[ListNode]:
        if not head:
            return None

        def delete(st, end):
            temp = end.next
            if temp:
                temp.prev = st
            end.next = None
            if st:
                st.next = temp
            return temp


        def checkAndGetNext(last):
            running = 0
            trav = last
            while trav:
                running += trav.val
                trav = trav.prev
                if running == 0:
                    return delete(trav, last)
            return last.next



        prev,trav = None,head
        running = 0
        # build prev
        while trav:
            trav.prev = prev
            running += trav.val
            prev = trav
            trav = trav.next
        if running == 0:
            return None

        trav = head
        while trav:
            trav = checkAndGetNext(trav)
            # print(trav)
            if trav and trav.prev == None:
                head = trav
        return head

        '''
        1 <- 2 <- -3 <- 3 <- 1

        
         '''

95