搜索二叉搜索树，使用计数器进行迭代

Question

I have a function I'm using that searches for a number in a binary search tree. I use an external int that can be incremented by the function with each iteration, since the function calls itself if the number is not the root.

I'm just learning binary search trees, but I know there's a better way of doing it that I'm overlooking... using an int method that returns the counter for example, but I can't figure it out...

edit: I know the number will definitely be in the BST, I just need to see how effective BST searching is vs. searching through an array for the same number.

// This external int can be incremented by the searchBST function 
// to keep track of iterations

int bcounter = 0;

// Search Binary Search Tree function

node* searchBST(node ** tree, int num){

bcounter++;

if(num < (*tree)->data) {
    searchBST(&((*tree)->left), num);
}
else 
    if(num > (*tree)->data) {
    searchBST(&((*tree)->right), num);
}
else 
    if(num == (*tree)->data) {
    return *tree;
}
}

Answer 1

Your counter is counting the depth of the found element, not the number of elements in the tree. If that is what you were looking for, then you're good.

If you want to find out the number of elements, you need to do something like

if (node is not null) {
  // count this node, it is not null
  count++;
  visit(node left);
  visit(node right);
}

it really doesn't matter if you visit left or right first, you just want to visit all the nodes of interest.