我实现了一个deleted函数，它从二叉树中删除一个特定的节点。num\u child（）返回一个节点的子节点数，get\u left（）返回左子节点数，依此类推。它适用于我尝试过的所有情况。我想知道代码是否正确？有没有什么方法可以简化代码？

def _delete(self):

        # if the node has 2 children
        if self.num_child() == 2:
            print("Cannot delete the node coz it has two child")

        # if the node has 1 child

        if self.num_child() == 1:
            #if left child is present
            if self.get_left() is not None:

                child=self.left
                #for root condition
                if self.parent is None:
                    self.left.parent = None
                    self.left= None

                else:
                    self.left.parent = self.parent
                    if self.data is self.parent.get_left():
                        self.parent.left = self.left
                    if self.data is self.parent.get_right():
                        self.parent.right = self.left

            #if right child is present
            if self.get_right() is not None:

                #for root condition
                if self.parent is None:
                    self.right.parent = None
                    self.right= None

                else:
                    self.right.parent = self.parent
                    if self.data is self.parent.get_left():
                        self.parent.left = self.right
                    if self.data is self.parent.get_right():
                        self.parent.right = self.right
            print(f"{self.data} has been deleted")
            return
        if self.num_child() == 0:
            if self.data is self.parent.get_left():
                self.parent.left = None
                self.parent= None

            elif self.data is self.parent.get_right():
                self.parent.right = None
                self.parent= None

            print(f"{self.data} has been deleted")