<p>You are given the head of a singly linked-list. The list can be represented as:</p>

<pre>
L<sub>0</sub> &rarr; L<sub>1</sub> &rarr; &hellip; &rarr; L<sub>n - 1</sub> &rarr; L<sub>n</sub>
</pre>

<p><em>Reorder the list to be on the following form:</em></p>

<pre>
L<sub>0</sub> &rarr; L<sub>n</sub> &rarr; L<sub>1</sub> &rarr; L<sub>n - 1</sub> &rarr; L<sub>2</sub> &rarr; L<sub>n - 2</sub> &rarr; &hellip;
</pre>

<p>You may not modify the values in the list&#39;s nodes. Only nodes themselves may be changed.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/04/reorder1linked-list.jpg" style="width: 422px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4]
<strong>Output:</strong> [1,4,2,3]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/03/09/reorder2-linked-list.jpg" style="width: 542px; height: 222px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5]
<strong>Output:</strong> [1,5,2,4,3]
</pre>

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

<ul>
	<li>The number of nodes in the list is in the range <code>[1, 5 * 10<sup>4</sup>]</code>.</li>
	<li><code>1 &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
from collections import deque
class Solution:
    def reorderList(self, head: Optional[ListNode]) -> None:

        queue = deque()
        current = head
        while current:
            queue.append(current)
            current = current.next
        
        new_head = ListNode(0)
        current = new_head
        left = True
        while queue:
            next = None
            if left:
                next = queue.popleft()
                left = False
            else:
                next = queue.pop()
                left = True
            next.next = None
            current.next = next
            current = current.next
        return new_head.next


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