In Griffiths, he defines a bound state to be that stationary state for which the total energy E is such that $E<V(\pm\infty)$. Let $\psi(x)$ is a stationary state satisfying $E<V(\pm\infty)$ and therefore a bound state.
On the other hand, a bouns state can also be defined as those stationary states which behave as $\psi(x)\rightarrow 0$ as $x\rightarrow \pm\infty$?
Is it possible to arrive at the second definition starting from the first or vice-versa?
Yes, it is of course possible. Is can be done rigorously, but if you like to have it simple, I will sketch a very nice and intuitive way.
It is convenient to understand Schroedinger's equation as an operator equation: $$\frac{p^2}{2m} = E_\mathrm{kin} = E_\mathrm{total} - V$$ Here $p^2$ is proportional to -$\frac{\mathrm d^2}{\mathrm dx^2}$, $V$ might depend on $x$, and $E_\mathrm{total}$ is just a number.
So you see, you get different differential equations for the two possible signs of $E_\mathrm{total} - V$. If it is positive, you get $\psi'' = -\psi$ and if it is negative you get $\psi'' = +\psi$. I threw away the constants like $\hbar$ and also pretended for simplicity, that the value of $V$ is constant. But this does not change the argument qualitatively if only the sign of $E-V$ does not change from a certain $x$ on. So my assumption is: there shall be anything happening for small $|x|$, but from a certain $|x|$ the potential shall get monotonic.
So there we are: for the first possibility for the sign, $e^{\pm \mathrm ix}$ are solutions, for the second $e^{\pm x}$. So the first possibility gives unbound states according to both your definitions, the second gives you bound states according to the first definition ($E<V$) and also bound states according to the second definition (the solution $e^x$ is impossible for positive $x$ and only $e^{-x}$ remains; and vice versa for negative $x$)
