Question

how would i do this? I am not sure when I would stop the bst search.

Answer 1

If each node of your tree has a field numLeft that tells you how many nodes there are in its left subtree (counting itself too), then you can do this in O(log N)

Just keep adding numLeft to a global result variable for each node whose value is less than x:

countLessThan(int x, node T)
    if T = null
        return
    if T.value >= x
        countLessThan(x, T.left) // T.left contains only numbers < T.value and T.right only numbers > T.value
    else
        globalResult += T.numLeft
        countLessThan(x, T.right)

This will only count the numbers. If you want to print them, you need to write a depth first traversal that will print a subtree given as parameter. You can find plenty of those online, so I won’t post that.