Assuming that n is a power of 2, 0 <= h1 <= h, T(0, h) = 0, and T(1, h) = h, an upper bound is 2nh by the following inductive proof.
Basis: T(0, h) <= 0 <= 2(0)h, and T(1, h) <= h <= 2(1)h.
Inductive step: T(n, h) = T(n/2, h1) + T(n/2, h - h1) + nh <= 2(n/2)h1 + 2(n/2)(h - h1) + nh <= 2nh, regardless of h1.