题目：给定一个单链表，其中的元素按升序排序，将其转换为高度平衡的二叉搜索树。本题中，一个高度平衡二叉树是指一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1。

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def sortedListToBST(self, head: ListNode) -> TreeNode:def rebuildTree(subhead):if not subhead:return Noneslow = subheadfast = subheadpre_next = ListNode(0)p = pre_nextwhile fast and fast.next:fast = fast.next.nextp.next = slowp = p.nextslow = slow.nextroot_val = slow.valroot = TreeNode(root_val)p.next = Noneroot.left = rebuildTree(pre_next.next)root.right = rebuildTree(slow.next)return rootreturn rebuildTree(head)执行结果：通过显示详情
执行用时 :152 ms, 在所有 python3 提交中击败了40.20%的用户
内存消耗 :16.1 MB, 在所有 python3 提交中击败了100.00%的用户

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def sortedListToBST(self, head: ListNode) -> TreeNode:nums = []p = headwhile p:nums.append(p.val)p = p.nextdef rebuildTree(left, right):if left >= right:return Noneroot_idx = (left+right) // 2root_val = nums[root_idx]root = TreeNode(root_val)root.left = rebuildTree(left, root_idx)root.right = rebuildTree(root_idx + 1, right)return rootreturn rebuildTree(0, len(nums))执行结果：通过显示详情
执行用时 :128 ms, 在所有 python3 提交中击败了95.75%的用户
内存消耗 :19.1 MB, 在所有 python3 提交中击败了75.97%的用户

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def sortedListToBST(self, head: ListNode) -> TreeNode:size = 0p = headwhile p:size += 1p = p.nextdef rebuildTree(l, r):nonlocal headif l >= r:return Nonemid = (l+r)//2left = rebuildTree(l, mid)root = TreeNode(head.val)root.left = lefthead = head.nextroot.right = rebuildTree(mid+1, r)return rootreturn rebuildTree(0 ,size)执行结果：通过显示详情
执行用时 :132 ms, 在所有 python3 提交中击败了91.67%的用户
内存消耗 :20.5 MB, 在所有 python3 提交中击败了5.19%的用户