The almost simple algebraic groups $G$ are classified by their root datum (defined here): in short, this is a quadruple $(X,X^{\ast},R,R^{\ast})$ with $X,X^{\ast}$ dual finitely generated free abelian groups and $R\subset X, R^{\ast}\subset X^{\ast}$ finite subsets (satisfying some conditions). Think of $X$ as the character group of some maximal torus in $G$ and $X$ a collection of roots (in particular, $R$ will define a root system once we tensor $X$ with $\mathbb{R}$). Then, $R^{\ast}\subset X^{\ast}$ are the corresponding co-objects (cocharacters, coroots). This datum characterises almost simple algebraic groups in the following sense: the type of $G$ is determined by the root system determined by $R\subset X$ (type ABCDEFG, as you've mentioned) and the extra information supplied by knowing the co-objects determines the weight lattice of this root system. The simply connected groups are those groups for which the weight lattice of the root system of $G$ is equal to $X$; this is the same as those groups of each type with the `largest' (finite) centre. The simple groups are those groups for which $\mathbb{Z}R = X$, that is, when the root lattice equals the character lattice.

Some nice, introductory notes are available here. Also, see Chevalley's Collected Works here.

I suppose this does not completely answer your question (I haven't given a list of the simply connected simple groups) but hopefully this will give you an idea of the general framework.

EDIT: I appear to have been too slow! As I was off looking for the required list @Jay Taylor has given a complete answer.

The almost simple algebraic groups $G$ are classified by their root datum (defined here): in short, this is a quadruple $(X,X^{\ast},R,R^{\ast})$ with $X,X^{\ast}$ dual finitely generated free abelian groups and $R\subset X, R^{\ast}\subset X^{\ast}$ finite subsets (satisfying some conditions). Think of $X$ as the character group of some maximal torus in $G$ and $X$ a collection of roots (in particular, $R$ will define a root system once we tensor $X$ with $\mathbb{R}$). Then, $R^{\ast}\subset X^{\ast}$ are the corresponding co-objects (cocharacters, coroots). This datum characterises almost simple algebraic groups in the following sense: the type of $G$ is determined by the root system determined by $R\subset X$ (type ABCDEFG, as you've mentioned) and the extra information supplied by knowing the co-objects determines the weight lattice of this root system. The simply connected groups are those groups for which the weight lattice of the root system of $G$ is equal to $X$; this is the same as those groups of each type with the `largest' (finite) centre. The simple groups are those groups for which $\mathbb{Z}R = X$, that is, when the root lattice equals the character lattice. Hence, the simple simply connected groups are those $G$ whose weight lattice is equal to its root lattice.

Some nice, introductory notes are available here. Also, see Chevalley's Collected Works here.

I suppose this does not completely answer your question (I haven't given a list of the simply connected simple groups) but hopefully this will give you an idea of the general framework.

EDIT: I appear to have been too slow! As I was off looking for the required list @Jay Taylor has given a complete answer.

The almost simple algebraic groups $G$ are classified by their root datum (defined here): in short, this is a quadruple $(X,X^{\ast},R,R^{\ast})$ with $X,X^{\ast}$ dual finitely generated free abelian groups and $R\subset X, R^{\ast}\subset X^{\ast}$ finite subsets (satisfying some conditions). Think of $X$ as the character group of some maximal torus in $G$ and $X$ a collection of roots (in particular, $R$ will define a root system once we tensor $X$ with $\mathbb{R}$). Then, $R^{\ast}\subset X^{\ast}$ are the corresponding co-objects (cocharacters, coroots). This datum characterises almost simple algebraic groups in the following sense: the type of $G$ is determined by the root system determined by $R\subset X$ (type ABCDEFG, as you've mentioned) and the extra information supplied by knowing the co-objects determines the weight lattice of this root system. The simply connected groups are those groups for which the weight lattice of the root system of $G$ is equal to $X$; this is the same as those groups of each type with the `largest' centre.

Some nice, introductory notes are available here. Also, see Chevalley's Collected Works here.

I suppose this does not completely answer your question (I haven't given a list of the simply connected groups) but hopefully this will give you an idea of the general framework.

The almost simple algebraic groups $G$ are classified by their root datum (defined here): in short, this is a quadruple $(X,X^{\ast},R,R^{\ast})$ with $X,X^{\ast}$ dual finitely generated free abelian groups and $R\subset X, R^{\ast}\subset X^{\ast}$ finite subsets (satisfying some conditions). Think of $X$ as the character group of some maximal torus in $G$ and $X$ a collection of roots (in particular, $R$ will define a root system once we tensor $X$ with $\mathbb{R}$). Then, $R^{\ast}\subset X^{\ast}$ are the corresponding co-objects (cocharacters, coroots). This datum characterises almost simple algebraic groups in the following sense: the type of $G$ is determined by the root system determined by $R\subset X$ (type ABCDEFG, as you've mentioned) and the extra information supplied by knowing the co-objects determines the weight lattice of this root system. The simply connected groups are those groups for which the weight lattice of the root system of $G$ is equal to $X$; this is the same as those groups of each type with the `largest' (finite) centre. The simple groups are those groups for which $\mathbb{Z}R = X$, that is, when the root lattice equals the character lattice.

Some nice, introductory notes are available here. Also, see Chevalley's Collected Works here.

I suppose this does not completely answer your question (I haven't given a list of the simply connected simple groups) but hopefully this will give you an idea of the general framework.

EDIT: I appear to have been too slow! As I was off looking for the required list @Jay Taylor has given a complete answer.