When you're asking about "exact" memory usage, do consider that all of those pointers may not be necessary. To see why, consider that the number of binary trees with $n$ nodes is $C_{2n}$, where:
$$C_i = \frac{1}{i+1} { 2i \choose i }$$
are the Catalan numbers. Using Stirling's approximation, we find:
$$\log C_{2n} = 2n - O(\log n)$$
So to represent a binary tree with $n$ nodes, it is sufficient to use $2n$ bits.
It's not too difficult to work how how to compress a static (i.e. non-updatable) binary search tree down to that size, though it is much harder to see how to retain $O(\log n)$ access time.
Incidentally, while different balanced binary tree variants are interesting from a theoretical perspective, the consistent message from decades of experimental algorithmics is that in practice, any balancing scheme is as good as any other. The purpose of balancing a binary search tree is to avoid degenerate behaviour, no more and no less. Stepanov also noted that if he'd designed the STL today, he might consider in-memory B-trees instead, because they use cache more efficiently.
As for hash tables, there is a similar analysis that you can do. If you are (say) storing $2^n$ integers in a hash table from the range $[0,2^m)$, and $2^n \ll 2^m$, then you can achieve this using close to:
$$\log {2^m \choose 2^n} \approx (n-m)2^n$$
bits.