Waves are generally classified as either transverse or longitudinal depending on the they way the propagated quantity is oriented with respect to the direction of propagation. Then what is a gravitational wave? It doesn't make sense to me that a disturbance in the curvature of spacetime has a "direction", so I would say they're neither, much like a wave packet in quantum mechanics.
2 Answers
Gravitational waves are transverse waves but they are not dipole transverse waves like most electromagnetic waves, they are quadrupole waves. They simultaneously squeeze and stretch matter in two perpendicular directions. Gravitational waves definitely propagate in a given direction but the effect that they have on matter is completely perpendicular to the direction of motion. Below is a picture of what the metric of a passing wave does to space (the wave traveling is perpendicular to the screen). If you imagine a free particle sitting at each grid intersection point, the particle would move sinusoidally right along with the grid:
This diagram is from this paper

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in the linearized limit of General relativity, as FrankH said, all propagating perturbations of the metric are transverse.
However, it must be noted that the full theory does allow for nonlinear longitudinal modes of propagation. According to the Petrov classification, such regions of longitudinal propagation are region III. There are usually not taken as true propagating modes since they fall of as the inverse cube of distance to the source, so they are interpreted as a kind of evanescent wave

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1Is this like in a waveguide for electromagnetic waves in which you can have longitudinal modes despite electromagnetic waves being generally transverse? – Ignacio Oct 28 '12 at 09:34
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1@Ignacio Since waveguides by definition don't propagate out in all directions, a better analogy would be to the near field of an EM transmitter. This regime has more complicated behavior, including the feedback into the transmitter that makes RFID technology possible. – Feb 09 '13 at 23:43
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1+1 especially for mention of the Petrov Classification. Schutz doesn't mention it: which is an oversight because, although not "practical" for gravitational wave astronomy, to at least know that other "near field" waves exist makes the whole subject of gravitational waves much less mysterious. – Selene Routley Sep 29 '15 at 13:10
and finally the traceless condition $$ H^{\mu}{; \mu} = 0 $$
yes? Then the other wave is the $H{\times}$ (again, H-cross ?) wave. Do you have a similiar diagram for this? Your diagram of the metric is far better than the usual one one finds of four test masses oscillating. Thank you. – Flint72 May 24 '14 at 15:29