如何返回按顺序遍历指定子树(指定节点的左子树或右子树)的数组

# 定义二叉树节点类
class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

# 返回按顺序遍历指定子树的数组
def traverse_subtree(root, subtree):
    if root is None:
        return []
    
    if subtree == "left":
        return traverse_subtree(root.left, subtree) + [root.val] + traverse_subtree(root.right, subtree)
    elif subtree == "right":
        return traverse_subtree(root.left, subtree) + traverse_subtree(root.right, subtree) + [root.val]
    else:
        return []

# 创建二叉树
root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
root.left.left = TreeNode(4)
root.left.right = TreeNode(5)
root.right.left = TreeNode(6)
root.right.right = TreeNode(7)

# 遍历左子树
left_subtree = traverse_subtree(root, "left")
print(left_subtree)  # 输出: [4, 2, 5, 1, 6, 3, 7]

# 遍历右子树
right_subtree = traverse_subtree(root, "right")
print(right_subtree)  # 输出: [2, 4, 5, 6, 7, 3, 1]