设计Twitter问题在Leetcode上我的错误在哪里？

设计Twitter问题在Leetcode上是一道经典的面试题。它的难度级别为中等（Medium），需要实现一个简单的社交媒体平台，具备发布推文、关注和查看推文等功能。

以下是完整的实现代码，假设你已经实现了以下三个类：User、Tweet和Twitter：

class User:
    def __init__(self, user_id):
        self.user_id = user_id
        self.followed = set()
        self.tweets = []

    def follow(self, user_id):
        self.followed.add(user_id)

    def unfollow(self, user_id):
        if user_id in self.followed:
            self.followed.remove(user_id)

    def post(self, tweet_id):
        self.tweets.append(tweet_id)

class Tweet:
    def __init__(self, tweet_id, user_id):
        self.tweet_id = tweet_id
        self.user_id = user_id
        self.timestamp = time.time()

class Twitter:
    def __init__(self):
        self.users = {}

    def postTweet(self, user_id, tweet_id):
        if user_id not in self.users:
            self.users[user_id] = User(user_id)
        self.users[user_id].post(tweet_id)

    def getNewsFeed(self, user_id):
        if user_id not in self.users:
            return []
        tweets = []
        for tweet in self.users[user_id].tweets:
            tweets.append(tweet)
        for followed in self.users[user_id].followed:
            if followed in self.users:
                for tweet in self.users[followed].tweets:
                    tweets.append(tweet)
        tweets = sorted(tweets, key=lambda x: x.timestamp, reverse=True)
        return [tweet.tweet_id for tweet in tweets[:10]]

    def follow(self, follower_id, followee_id):
        if follower_id not in self.users:
            self.users[follower_id] = User(follower_id)
        self.users[follower_id].follow(followee_id)

    def unfollow(self, follower_id, followee_id):
        if follower_id not in self.users:
            return
        self.users[follower_id].unfollow