The difference lies in spatial infinity: AdS like spacetimes have timelike infinity while Schwarzschild spacetime has a lightlike infinity.
In general the prescription to build <PERSON> diagrams of spherically symmetric spacetimes goes as:
1) Define the tortoise coordinate except for the constant;
2) Define null coordinates in the 'usual way';
3) Define new null coordinates in order to avoid the coordinate singularity in the event horizon;
4) Reparametrize the null coordinate in order to put it in a finite diagram.
In the case of AdS-Schwarzschild spacetime it is a little bit of hard work:
1) We have a metric of type $g_{\mu \nu }= (-f, f^{-1},r^2,r^2\sin^2 \theta )$ and $ f = 1- \frac{2m}{r} + c^2r^2$, $c$ meaning the inverse of the AdS curvature radio. Now, as long as $f$ has always a real positive and two imaginary roots, we have \begin{eqnarray} r_* = \int \frac{1}{f}dr = \frac{(2r_+^2 + 4p)\arctan \left(\frac{r_++2r}{\sqrt{4p-r_+^2}}\right) + r_+ \sqrt{4p-r_+^2} \ln \left( \frac{(r-r_+)^2}{r^2+r_+r+p} \right)}{2c^2(2r_+^2+p)\sqrt{4p-r_+^2}} + constant \end{eqnarray} in which $r_+$ is the event horizon ($f(r_+)=0$) and $p=r_+^2+\frac{1}{c^2}= \frac{2m}{c^2r_+}$. (Assume: $constant \equiv C$).
2) The usual null-coordinates, $u$ and $v$ written as \begin{eqnarray} du=dt-dr_ \ dv=dt+dr_ \end{eqnarray} lead the metric to \begin{eqnarray} ds^2 = -f du dv + r^2 d \Omega^2 \end{eqnarray} which is still singular in $r=r_+$ as long as $f=0$ there.
3) Defining now \begin{eqnarray} \tilde{U}=-e^{-\eta u}\ \tilde{V}=e^{\eta v}, \end{eqnarray} we have \begin{eqnarray} ds^2 = \frac{-f}{\eta^2}e^{-2\eta r_*}d \tilde{U} d\tilde{V} +r^2 d\Omega^2 \end{eqnarray} (being $\tilde{U}$ and $\tilde{V}$, also null). 578
If $\eta = \frac{c^2(2r_+^2+p)}{2r_+}$, the coordinate singularity is avoided.
With these new coordinates, and the proper choice of $C$ (in blocks) we have, \begin{eqnarray} \lim_{r\rightarrow \infty} \tilde{U} \tilde{V} = -1 \ \lim_{r\rightarrow r_+} \tilde{U} \tilde{V} = 0 \ \lim_{r\rightarrow 0} \tilde{U} \tilde{V} = 1. \end{eqnarray}
4) Finally, define \begin{eqnarray} U = \arctan \tilde{U}\ V = \arctan \tilde{V} \end{eqnarray}
in which case, the maximal spacetime is placed in a square of size $\pi / \sqrt{2}$. To see this, we have to take the 3 limits above, considering the coordinates $U$ and $V$ as in the file bellow. 804