tl;dr -- See the question title
I quote from this Wikipedia article on the depletion region:
By definition, the N-type semiconductor has an excess of free electrons (in the conduction band) compared to the P-type semiconductor, and the P-type has an excess of holes (in the valence band) compared to the N-type. Therefore, when N-doped and P-doped semiconductors are placed together to form a junction, free electrons in the N-side conduction band migrate (diffuse) into the P-side conduction band, and holes in the P-side valence band migrate into the N-side valence band.
Pardon my naivete (if the phrasing or question suggest any), Physics is not really my of area expertise; but my understanding prior to this had been that the diffusion of holes in the p-type region to the n-type region was really a consequence of the n-type conduction band electrons "filling" the former positions of the p-type holes, and conversely, n-type valence band electrons equivalent in quantity to the now diffused electrons occupying (somehow) the evacuated positions of the diffused electrons in the conduction band--thus "creating" the appropriate number of "diffused holes" in the n-type region.
However this would also suggest that the conduction band of the n-type region is continually replenished--which is apparently wrong by this answer:
Higher energies are occupied on the left than the right, and what happens? Answer: some of the higher-energy electrons on the left "fall into" the lower-energy states on the right. Yes, this leaves a + charge on the n-type side and a − charge on the p-type side with an associated voltage. [...] The result: on the right hand side all of your valence band is filled in; on your left hand side none of your conduction band remains, and you've got a "depletion region".
What I'm failing to understand is how the p-type holes would then diffuse to the n-type region if (1) the n-type electrons diffuse to the p-type conduction band and (2) the conduction band in the n-type material is left devoid of electrons. Would appreciate your responses.