As is well-known, the representations of $SU(n)$ are labelled by Young diagrams. Moreover, there exists a canonical basis of each representation labelled by all the possible tableaux of the diagram.
What happens for the B and C series, i.e. the compact symplectic groups and the special orthogonal groups? My guess is that the representations are still labelled by Young diagrams, but with a different basis.
This is explained very well in Fulton and Harris. You need to look at Chapter 17.3 for the symplectic case and Chapter 19.5 for the orthogonal case.
Let me quickly sketch the symplectic case, $\mathfrak{sp}_{2n} \mathbb C$. Take to $V = \mathbb C^{2n}$ to be the representation defined by the $\mathfrak{sp}_{2n} \mathbb C$ matrices themselves. Given a Young diagram $\lambda$ with at most $2n$ rows and $2n$ columns, you can act with the associated Schur functor to get a new representation $\mathbb S_\lambda V$, just like the $\mathfrak{sl}_n\mathbb C$ case that you're familiar with.
However, there is a novelty in the $\mathfrak{sp}_{2n} \mathbb C$ case, in that $\mathbb S_\lambda V$ in itself is not the irreducible representation you're looking for. In fact, the irreducible representation you want is a subrepresentation of $\mathbb S_\lambda V$. To get it, remember that $\mathbb S_\lambda V$ is naturally a subspace of $V^{\otimes d}$, where $d$ is the number of boxes in your Young diagram. You consider all possible contractions $V^{\otimes d} \to V^{\otimes (d-2)}$ given by $$ v_1 \otimes \dots \otimes v_d \to Q(v_p, v_q) v_1 \otimes \dots \otimes \hat{v_p} \otimes \dots \otimes \hat{v_q} \otimes \dots \otimes v_d,$$ where $Q$ is the natural symplectic form on $\mathfrak{sp}_{2n}\mathbb C$ and $1 \leq p < q \leq d$. The irreducible representation you're looking for is the intersection of $\mathbb S_\lambda V$ with the kernels of all such contractions for all $p$ and $q$.
Actually that still isn't quite right: although you can draw Young diagrams with up to $2n$ rows and the Schur functor $\mathbb S_\lambda \mathbb C^{2n}$ is perfectly well defined, it turns out that this construction of the irreducible representations only works if the Young diagram has $n$ rows or fewer; if you have more than $n$ rows, this construction just gives the zero vector space.
As in the $\mathfrak{sl}_n\mathbb C$ case, the number of boxes in each row of the Young diagram are the coefficients of the highest weight of the corresponding irrep, written with respect to a basis of simple weights. Since irreps are in one-to-one correspondence with possible highest weight vectors, this construction gives all of the irreps of $\mathfrak{sp}_{2n} \mathbb C$.