When a quantity is within 2 units of 0, its distance from 0 is at most 2. Thus $|t + 3| \leq 2$ is equivalent to $-2 \leq t + 3 \leq 2$, which leads to $-5 \leq t \leq -1$. Alternatively, we can obtain the same result by considering two cases, $t + 3 \geq 0$ and $t + 3 < 0$: when $t + 3 \geq 0$, we have $|t + 3| = t + 3 \leq 2$, which gives $-5 \leq t \leq -1$; and when $t + 3 < 0$, we have $|t + 3| = -(t + 3) \leq 2$, which is equivalent to $-5 \leq t \leq -1$. Thus, the minimum and maximum values of $t$ are $\boxed{-5}$ and $\boxed{-1}$, respectively.
