Night light, schematic and functioning
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. Redraw of the OP's reverse engineering.


If your circuit is correct then we can see three voltage dropping resistors, R3, 4 and 5, and a half-wave rectifier, D4. At 240 V the current through the resistors will be \$ I = \frac {240}{8k2 + 8k2+8k2} = 10~mA \$ but with the rectifier it will average half that.
It's not clear from your schematic but I suspect that R2 is the light sensor - an LDR. When light is sensed the resistance will drop and Q2 will turn on. This will "shunt" the DC on C1 to ground and turn the LEDs off. This will give comfort to the user giving the impression that the unit is not wasting power when, in fact, it is running constant power whether it's on or off. It would make no difference to power consumption if R1, 2 and Q2 were omitted!
Power dissipated in each of the resistors will be \$P = I^2R = (5m)^2 \cdot 8k2 = 205~mW \$ which may be a bit on the high side for those SMD resistors.