As shubham notes, the Schroedinger equation is indeed a Helmholtz equation in 1D (your case), but the general form of the equation is:
$$
- \frac{\hbar^2}{2m} \nabla^2 \Psi \left( \mathbf{r}, t \right) + \operatorname{V \left( \mathbf{r},t \right)} \Psi = i \hbar \frac{\partial \Psi \left(\mathbf{r}, t\right)}{\partial t}
$$
This is a diffusion equation in which the diffusion equation is a constant. For comparison, we have (if $V = 0$):
$$
D\nabla^2 \Psi = \frac{\partial \Psi}{\partial t}
\quad
\implies D = \frac{i \hbar}{2 m}
$$
However, if we have a static case, then the term on the right hand side reduces to a Helmholtz equation (if we have $V = 0$ everywhere in our space):
$$
\nabla^2 \Psi + \frac{2mE}{\hbar^2} \Psi
$$
where you can infer that
$$
\left|\mathbf{k}\right|^2 = \frac{2mE}{\hbar^2}
$$
Which, as you see is a general result, which holds in 3D as well.