quantum mechanics - Difference for boundary condition, particle in a box

When solving the simple problem of a free particle in a box of volume $V = L^3$, we can impose either periodic boundary conditions $\psi(0) = \psi(L)$ and $\psi '(0)= \psi'(L)$ either strict boundary condition $\psi(0)=0$ and $\psi(L)=0$.

Now, when looking for the eigenstates of the Hamiltonian, one can easily solve the problem with the separation of variables method, and impose the boundary conditions.

For the case of periodic boundaries, one has that the eigenvalue in each direction $(x,y,z)$ of space must be $\lambda_k = 2\pi k/L$, while in the strict boundaries you find $\lambda_k = \pi k/L$.

Now, in every textbook I could find, they say that in the first case we have that $k\in \mathbb{Z}$ and in the second $k \in \mathbb{N}$.

For the case $k = 0$ I can understand the difference, but I do not understand why negatives $k$ in the case of boundary condition must be discarded.

Edit : (Sorry for bumping) I understand why "physically" the solutions with periodic boundary conditions differ from the one with strict boundary conditions, and you're answer is perfectly sensible in that way.

However, surely there must be a way of showing (mathematically, I mean) that in the case of strict conditions choosing $k \in \mathbb{Z}$ yields "duplicate" solutions? Could someone offer some insight concerning this proof?