Driving bicolor LED from one I/O via 74HC595 shift register

Why such a complicated diagram.

R1, D1, R2, D2 can be replaced with ONE 220ohm resistor.
R3, U1A can be removed, and the HC595 output can be connected directly to the above 220ohm resistor and the input of U1B.

The 74HC14 has only six inverters, so you need even more chips.
Leo..