<p>Given the <code>head</code> of a linked list, return <em>the list after sorting it in <strong>ascending order</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/14/sort_list_1.jpg" style="width: 450px; height: 194px;" />
<pre>
<strong>Input:</strong> head = [4,2,1,3]
<strong>Output:</strong> [1,2,3,4]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/14/sort_list_2.jpg" style="width: 550px; height: 184px;" />
<pre>
<strong>Input:</strong> head = [-1,5,3,4,0]
<strong>Output:</strong> [-1,0,3,4,5]
</pre>

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

<pre>
<strong>Input:</strong> head = []
<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, 5 * 10<sup>4</sup>]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Can you sort the linked list in <code>O(n logn)</code> time and <code>O(1)</code> memory (i.e. constant space)?</p>

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

            # split nodes into two halfs
            p, s, f = None, head, head
            while f and f.next:
                p = s
                s = s.next
                f = f.next.next

            p.next = None
            leftList = self.sortList(head)
            rightList = self.sortList(s)

            # merging
            tail = dummy = TreeNode(0) 
            while leftList and rightList:
                if leftList.val <= rightList.val:
                    new = TreeNode(leftList.val)
                    tail.next = new
                    tail = tail.next
                    leftList = leftList.next
                else:
                    new = TreeNode(rightList.val)
                    tail.next = new
                    tail = tail.next
                    rightList = rightList.next

            if leftList:
                tail.next = leftList
            
            if rightList:
                tail.next = rightList

            return dummy.next    


        # return merge(head)    
34700