用python遍历二叉搜索树 ，然后将其转换为循环的双链表，最后按正反序输出双链表中的所有元素

好的，下面是Python代码实现：

# 定义二叉树节点类
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None

# 将二叉搜索树转换为循环的双链表
def tree_to_list(root):
    if not root:
        return None
    stack = []
    cur = root
    pre = None
    head = None
    while cur or stack:
        while cur:
            stack.append(cur)
            cur = cur.left
        cur = stack.pop()
        if not head:
            head = cur
        if pre:
            pre.right = cur
            cur.left = pre
        pre = cur
        cur = cur.right
    head.left = pre
    pre.right = head
    return head

# 正序输出双向链表所有元素
def print_list(head):
    cur = head
    while cur.right != head:
        print(cur.val, end=' ')
        cur = cur.right
    print(cur.val)

# 反序输出双向链表所有元素
def print_list_reverse(head):
    cur = head.left
    while cur != head:
        print(cur.val, end=' ')
        cur = cur.left
    print(cur.val)

if __name__ == '__main__':
    # 构造二叉搜索树
    root = TreeNode(4)
    root.left = TreeNode(2)
    root.right = TreeNode(6)
    root.left.left = TreeNode(1)
    root.left.right = TreeNode(3)
    root.right.left = TreeNode(5)
    root.right.right = TreeNode(7)

    # 将二叉搜索树转换为循环的双链表
    head = tree_to_list(root)

    # 输出双向链表中所有元素
    print('正序输出双向链表：')
    print_list(head)

    print('反序输出双向链表：')
    print_list_reverse(head)

输出结果如下：

正序输出双向链表：
1 2 3 4 5 6 7
反序输出双向链表：
7 6 5 4 3 2 1