There are a couple of problems in your code. But let's go straight to the hurting point.
What's the main problem ?
From your code, it is obvious that tree1Ptr is a pointer. In principle, it should point to a node of the tree, which has two pointers, one to the left node and one to the right node.
So somewhere in your code, you should have:
tree1Ptr = new Node; // or whatever the type of your node is called
However, in your inner loop, you are just using it as if it was an array:
for (int i = 0; i < 10000; i++)
{
int random = rand() % 20000;
tree1Ptr->add(random);
for (int j = 0; j < i; j++) {
if (tree1Ptr[j]==random) //<============ OUCH !!
i--;
}
}
The compiler won't complain, because it's valid syntax: you can use array indexing on a pointer. But it's up to you to ensure that you don not go out of bounds (so here, that j remains <1).
Other remarks
By the way, in the inner loop, you just want to say that you have to retry if the number is found. You can break the inner loop if the number is already found, in order not to continue.
You should also seed your random number generator, to avoid running the program always with the same sequence.
How to solve it ?
You really need to deepen your understanding of BST. Navigating through the node requires make comparison with the value in the current node, and depending on the result, iterate continuing either with the left or the right pointer, not using indexing. But it would be too long to explain here. So may be you should look for a tutorial, like this one
