PIC18 how to explain : you write bit 1 by making TRIS input
The idea behind this is the following:


Setting the TRIS bit (making it 1) will make the corresponding pin an input.

A typical extract from a PIC datasheet:


  Setting a TRISA bit (= 1) will make the corresponding PORTA pin an input (i.e., put the corresponding output driver in a Hi-Impedance mode). Clearing a TRISA bit (= 0) will make the corresponding PORTA pin an output (i.e., put the contents of the output latch on the selected pin).

That the pin is in high impedance mode, means (among other things) that the pin is not driven by the chip itself. Therefore, it takes any state that is defined by the external circuit (the circuit around the PIC).
The 1-wire bus uses pull-up resistors. You can see that on the wikipedia article:


  


As you can see, the default state is high. That means that the bus uses a pull-up resistor to make it's default state high.

That's also written on the wikipedia:


  The Dallas 1-Wire network is physically implemented as an open drain master device connected to one or more open drain slaves. A single pull-up resistor is common to all devices and acts to pull the bus up to 3 or 5 volts, and may provide power to the slave devices. Communication occurs when a master or slave asserts the bus low, i.e. connects the pull up resistor to ground through its output MOSFET. Specific 1-Wire driver and bridge chips are also available. Data rates of 16.3 kbit/s can be achieved. There is also an overdrive mode which speeds up the communication by a factor of 10.