No, but you can be forgiven for thinking that with an NPN junction of a BJT looking terribly much like two PN junctions of two diodes wired together.

So let me give the physics backstory here. Both N-type and P-type doped semiconductors are conductors, and they're both electrically neutral -- but in one of them (N-type) we've added atoms which have "too many" electrons in their outer shell (compared to the semi) and those electrons form conduction states shared among all of the atoms; and in the other (P-type) we've added atoms which have "too few" electrons in their outer shell, and then these electron-holes form conduction states shared among all of the atoms.

We join these two at a PN-junction, and then something very "obvious" happens: the electrons fall into the holes. This is actually maybe not as straightforward as it sounds because we say that there is no "bias voltage" across these two but there is certainly an internal electric field, as the P-type side has more electrons than makes it neutral and the N-type side has fewer. What's actually happening is that what we call "voltage" is generally the "chemical potential" of the electrons which is normally dominated by electrical potential, but in this case we see a competing effect from diffusion, the excess electrons from the N-type dopants have diffused to the P-type dopants and thus both conduction bands evaporate in this "depletion region", where the diffusion tendency is balanced out by the electric field that points from the N-type side to the P-type side.

Now we put a "forward bias" on it, this means raising the P-type voltage relative to the N-type side, so we are putting more holes into the P-type side, more electrons into the N-type side. And they attract each other across this divide, but as they try to connect they get overwhelmed by the electric field between the two, and "roll back" to their side. This happens until the voltage bias is exactly large enough to cancel out the electric potential difference of the boundary: then electrons in the N-type side successfully can fly a little distance into the P-type side and holes from the P-type side can fly a little distance into the N-type side, allowing the local electrons to "fall into" the holes: and a net current can flow again! This voltage difference is actually a material property for the semiconductor in general; so for Silicon this "diode drop" is 0.7 volts; when forward-biased the diode appears to be something like an 0.7V battery wired against the current, when reverse-biased the diode appears to be a circuit break, and in the middle you might have some sort of interesting linear and nonlinear regions.

If you just connect two diodes together backwards, one or the other of them is going to be reverse biased unless you inject or extract current out of the region between them -- the only flows possible are "outside to center" and "center to outside" until you get to the breakdown regions of the diodes.

Okay now that we're clear on diodes let's talk about the NPN in a typical bipolar junction transistor. The key here is that the P region is actually very small. Usually we call one of the N regions (which is usually more heavily doped than the other) the "emitter" and the other the "collector". The usual recipe for the voltages in normal operation is,

Vb > Vc > Ve

so that electrons, wanting to flow from lower to higher voltages, really want to move from the emitter to the collector and most of all would like to move into the base. In addition, this Vb - Ve is generally about the same magnitude as a standard diode drop, so you will see a little diode symbol in the BJT circuit diagram symbol, which helps to remember this.

Since Vb - Ve is a typical diode drop, we see that holes in the base have no problem with combining with electrons in the emitter and current flows freely between these two. On the other hand we see that the depletion region between the base and the collector has actually grown.

This is where the thinness of the base is really important. The depletion region isn't really a place where electrons can't move, it's just a place where electrons coming from the N side find themselves repelled by an electric field from excess electrons on the P side. But we just said, the electrons coming in from the emitter are no longer being repelled by the excess electrons in the base; they're penetrating into the P side. Well, once that happens, the electrons from the emitter might also get far enough that they feel these electrons "behind them" pushing them into the collector, where they are likewise attracted by the unbalanced positive nuclei on the other side. So it's kind of a rollercoaster ride where electrons with high enough energies can fly into the base and then find themselves dumped into the collector.

This causes an electron current from the emitter to the collector which tends to be proportional to the electron current from the emitter to the base -- and the proportionality constant is usually governed by how thin this interface is, since it requires the electron to diffuse far enough across the base for the residual forces to push it into the collector. For modern BJTs this proportionality constant can be in the hundreds or thousands: for each hole that makes its way into the emitter to recombine, hundreds or thousands of electrons could be coaxed past this point-of-no-return where they fall into the collector.

The mental model you want here is that there is this base-emitter diode which, if it's reverse-biased, you usually see nothing (unless you totally reverse the BJT and pretend the emitter is the collector and vice versa --- usually the aforementioned current gain is much smaller) -- and you have an ideal current source connected pointing from the collector to the emitter, and the magnitude of that current source is given by this proportionality factor with the comparatively weak current going from the base into the emitter.