C中的BST搜索功能

Question

I have a problem with search function there.. I searched a lot of examples with returning values, but i just want to write it in void function and this function just have to print "Exist" or "Not exist".. Any help or hint how to do it? In C ofc.

struct bin_tree {
int data;
struct bin_tree * right, * left;
};

typedef struct bin_tree node;

void insert(node ** tree, int val)
{
    node *temp = NULL;
    if(!(*tree))
    {
        temp = (node *)malloc(sizeof(node));
        temp->left = temp->right = NULL;
        temp->data = val;
        *tree = temp;
    return;
    }
    if(val < (*tree)->data)
    {
        insert(&(*tree)->left, val);
    }
    else if(val > (*tree)->data)
    {
        insert(&(*tree)->right, val);
    }
}
void print_preorder(node * tree)
{
     if (tree)
     {
          printf("%d\n",tree->data);
          print_preorder(tree->left);
          print_preorder(tree->right);
     }
}
node* search(node ** tree, int val)
{
    if(!(*tree))
    {
        printf("Tree doesnt exist");
    }
    if(val < (*tree)->data)
    {
        search(&((*tree)->left), val);
    }
    else if(val > (*tree)->data)
    {
        search(&((*tree)->right), val);
    }
    else if(val == (*tree)->data)
    {
        printf("%d exist", val);
    }
    printf("%d not exist", val);
}
int main()
{
    node *root;
    node *tmp;
    int x, b;
    root = NULL;
    int warunek;
    do
    {
        printf("\nMENU\n");
        printf("1. Add[1]\n2. Preorder\n3. Search\n0. Exit[0]\n");
        scanf("%d", &warunek);
        switch(warunek)
        {
            case 1:
            {
                printf("Give a number: ");
                scanf("%d", &x);
                insert(&root, x);
            }
            break;
            case 2: print_preorder(root);
            break;
            case 3:
            {
                printf("Give a number to search: ");
                scanf("%d", &b);
                search(&root, b);
            }
            break;
            default: printf("Bad value, write 1, 2 or 3");
        }
    }
    while(warunek != 0);
    return 0;
}

Answer 1

Assuming all of the other functions correct I am just focusing on the function

node* search(node ** tree, int val)

You need to change the return type of the function to void as you are not returning any value -

void search(node ** tree, int val)

Now for the search functions you have checked whether *tree is NULL or not.
Note that this condition will occur in two cases -

• The tree passed to the function itself is empty.
• The element you are searching itself is not present.

The first case can be checked in the main function itself, before passing the tree root to the function. So you need to change your if part as -

if(!(*tree))
{
    printf("%d not exist", val);
}

In main function you need to make the following changes -

if(root!=NULL){
  printf("Give a number to search: ");
  scanf("%d", &b);
  search(&root, b);
 }
  else {
      printf("No search can be performed as the tree has no nodes.");
  }

Note that as mentioned in the comments the search function does not require a pointer to pointer to the root. But here I have just suggested the necessary minimal changes in your code. More efficient implementation exist.

Answer 2

Your search function is basicallyokay, but ...

• ... you must make all four possible outcomes — null node, number is equal, smaller or greater than the current node exclusive. That is,you have an if ... else if ... else if ... else ... chain.
• ... since you don't want to return anything, the return type of your function must ve void .
• ... you do not modify the contents of the tree, so it isn't necessary to pass in a pointer to the root pointer. That's only needed when you modify the tree structure, ie add, remove or reorder nodes.

So:

void search(const node *tree, int val)
{
    if (!tree) {
        printf("%d doesn't exist.\n", val);
    } else if (val < tree->data) {
        search(tree->left, val);
    } else if (val > tree->data) {
        search(tree->right, val);
    } else {
        printf("%d exists.\n", val);
    }
}

because you follow only one path through the tree, you can also make this function iterative:

int search(const node *tree, int val)
{
    while (tree) {
        if (val == tree->data) {
            printf("%d exists.\n", val);
            return 1;
        }

        if (val < tree->data) {
            tree = tree->left;
        } else {
            tree = tree->right;
        }
    }

    printf("%d doesn't exist.\n", val);
    return 0;
}

Thistime, I've returned a truth value. In my opinion, it is better to have the function not print anything, but print the outcome from the calling code.