// 打印前 n 个出现次数最多的单词 void print_top_n(struct tnode** top_n, int n) { printf("Top %d words:\n", n); int i,j; for ( i = 0; i < n; i++) { if (top_n[i]->num == 0) { break; } printf("%d: ", i + 1); struct tnode* cur = top_n[i]; char word[200]={'\0'}; int len = 0; // 从 Trie 树中回溯到根节点，构造出单词 while (cur != NULL) { word[len++] = 'a' + (cur - cur->ptr[0]); cur = cur->parent; } for ( j = len - 2; j >= 0; j--) { printf("%c", word[j]); } printf(" (%d occurrences)\n", top_n[i]->num); } }优化这段代码

这段代码的主要瓶颈在于从 Trie 树中回溯到根节点，构造出单词的操作。可以使用一个字符数组来保存当前的单词，避免反复申请和释放内存。同时，还可以将从 Trie 树中回溯到根节点的操作改为使用指针数组进行，提高代码的可读性和执行效率。优化后的代码如下：

void print_top_n(struct tnode** top_n, int n) {
    printf("Top %d words:\n", n);
    for (int i = 0; i < n && top_n[i]->num > 0; ++i) {
        printf("%d: ", i + 1);
        struct tnode* cur = top_n[i];
        char word[200];
        int len = 0;
        // 从 Trie 树中回溯到根节点，构造出单词
        while (cur->parent != NULL) {
            word[len++] = 'a' + (cur - cur->ptr[0]);
            cur = cur->parent;
        }
        // 将单词翻转回来
        for (int j = len - 1; j >= 0; --j) {
            printf("%c", word[j]);
        }
        printf(" (%d occurrences)\n", top_n[i]->num);
    }
}