Recovery of transistor wired as diode
This should actually be only a comment
CircuitLab transient analyzer shows that you are right. There's no such thing as storage time - not at least when the transistor is Cicuit Lab's version of 2N3904.  Here's a circuit with 1N4148.
Signal V2 is bipolar square wave 5V peak:


The diode conducts to reverse direction about 10 nanoseconds. Transistor turns off substantially faster with the same signal and load:


The reverse conduction starts to decay as soon as the polarity is reversed, there's no full reverse conduction period (= no storage time). Schottky-diodes have the same property. Except they have about 50% lower voltage drop.
Transistor's BE junction unfortunately stands only a few volts of reverse voltage, so this may be not so good in high voltage circuits.
Not asked: The output voltage to the load seems to be about -5,7V when V2 jumps to -5V. That can look surprising. It's not. The carrier cloud in the semiconductor is like a charged capacitor which has minus pole towards the load. When V2 jumps to negative there's -0,7V in series with V2.
The peak which occurs when V2 jumps to +5V is the same to another direction. The semiconductor capacitances have got bigger voltages during the minus polarity period of V2. Transistor seems to have much bigger capacitance than diode 1N4148,