Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.

链表转换为BST，找到最中间的node设置为树的root，然后截断（设null），对分离的两个链表再应用这个函数设置为上一级root的left child & right child.

class Solution:    def sortedListToBST(self, head):        if not head:            return None        if not head.next:            return TreeNode(head.val)        a, b = head, head.next        while b.next and b.next.next:            a = a.next            b = b.next.next        root = a.next #find the mid node        new_root = TreeNode(root.val) #set it to the root of the tree        a.next = None #the mid node was set to null, LL -> two LLs        new_root.left = self.sortedListToBST(head) #recall the function to the two LLs        new_root.right = self.sortedListToBST(root.next) #to construct the children        return new_root