No. In fact one shows by a simple “ODE‐continuity’’ argument that you cannot have a C^2 Lorentzian metric all of whose time– and space–like geodesics are complete but which admits an inextendible null geodesic in finite affine parameter. Equivalently

timelike & spacelike geodesic completeness ⇒ null geodesic completeness.

The heuristic is this. If γ0(λ) is an inextendible null geodesic ending at λ=λ*<∞, choose at the starting point p = γ0(0) a one–parameter family of timelike unit–vectors vε converging to the null tangent \dotγ0(0). By standard existence–and–uniqueness for geodesics and continuous dependence on initial data, the timelike geodesics γε(λ) with γε′(0)=vε exist at least as long as γ0 does, and in fact must run arbitrarily close to γ0(λ) for λ≲λ*. If all timelike geodesics were complete, those γε would extend past λ*, and by continuity γ0 would as well. Hence γ0 could not really have “hit’’ a boundary at λ*; it would extend beyond it.

A more rigorous statement can be found e.g. in Beem–Ehrlich–Easley
“Global Lorentzian Geometry,” Thm. 12.21. There it is proved that any smooth Lorentz‐metric which is complete on all non‐null geodesics is automatically complete on null ones.

So one cannot fill in the “missing’’ sixth case. In all smooth Lorentzian manifolds, if every timelike and every spacelike geodesic runs off to infinite parameter, so must every null geodesic.