C中二叉树的迭代器方法 [英] Iterator Method for Binary Trees in C

问题描述

我收到了以下说明：



向库中添加一个新方法，将函数应用于树中的每个值。 （这是树的迭代器方法。）

I've been given the following instruction:

Add a new method to the library that applies a function to every value in the tree. (This is an iterator method for trees.)

int bst_char_iterate(bst_char *t, char (*fun)(char item))

注意：只有传递给它的函数是单调的，才能通过此方法保留BST属性。每当

x< = y（箭头）f（x）< = f（y）时，函数f都是单调的。



Bst_char * t等同于指向树的指针，* fun等同于指向函数调用的指针，item是调用函数的值。



我设法得到了这个：

NB: The BST property will only be preserved by this method if the function passed to it is monotonic. A function f is monotonic whenever
x <= y (arrow) f(x) <= f(y).

"Bst_char *t" equating to a pointer to the tree, "*fun" equating to a pointer to a function call and "item" being the value of which the function is called upon.

I've managed to come up with this:

int bst_char_iterate(bst_char *t, char (*fun)(char item)) {
    assert(t!=NULL);
    struct node * p = t->root;
    if (p!=NULL) {
       p->item = fun(p->item);
       p->left->item = bst_char_iterate(t,fun(p->left));
       p->right->item = bst_char_iterate(t,fun(p->right));
    } else {
       return 0;
    }
}

这是无法编译的，有很多错误（例如来自没有指针的指针）演员）。



请帮忙！



编辑：我已设法让它编译这段代码：

This cannot compile, with quite a few errors (such as "from pointer without a cast").

Please help!

I have managed to get it to compile with this code:

int bst_iterate(struct node *p, char (*fun)(char item)) {
    if (p!=NULL) {
       p->item = fun(p->item);
       bst_iterate(p->left,fun);
       bst_iterate(p->right,fun);
    } else {
       return 0;
    }
}

int bst_char_iterate(bst_char *t, char (*fun)(char item)) {
    assert(t!=NULL);
    bst_iterate(t->root,(fun));
}

但我不知道如何在测试程序中实现它。



我试过：

But I've no idea how to implement it in the test program.

I've tried:

bst_char_iterate(t, fun);
bst_char_iterate(t, *fun);
bst_char_iterate(t->root, fun);
bst_char_iterate(t, fun(c));

推荐答案

Quote：

但是我不知道如何在测试程序中实现它。

But I've no idea how to implement it in the test program.

保持简单：

Keeping it simple:

// the dumbest monotonic function...
char  my_fun(char c)
{
  return c;
}


int main()
{
  bst_char * ptree;
  // ...
  // ... create the tree, make ptree pointing to it
  // ...
  bst_char_iterate(ptree, my_fun);
  // ...
  // ... free the tree
  // ...
  return 0;
}

bst_char_iterate(t, fun);         // what type is t?
bst_char_iterate(t, *fun);        // *fun is incorrect, use fun
bst_char_iterate(t->root, fun);   // what type is root?
bst_char_iterate(t, fun(c));      // fun(c) is wrong, use fun

在上述所有情况下，什么是有趣以及定义在哪里？



我在我的第一条评论中说，你需要显示编译器产生的确切错误信息。

And in all the above cases what is fun and where is it defined?

As I said in my first comment, you need to show the exact error message the compiler produces.

这篇关于C中二叉树的迭代器方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！