So for example our tree returns: '110', it should mean this byte:
'00000110' right?
Wrong. You should have a byte buffer of bits into which you write your bits. Write the three bits 110 into the byte. (You will need to decide on a convention for bit ordering in the byte.) You still have five unused bits in the byte, so there it sits. Now you write 10 into the buffer. The byte buffer now has 11010, and three unused bits. So still it sits. Now you try to write 111011 into the byte buffer. The first three bits go into the byte buffer, giving you 11010111. You now have filled the buffer, so only now do you write out your byte to the file. You are left with 011. You clear your byte buffer of bits since you wrote it out, and put in the remaining 011 from your last code. Your byte buffer now has three bits in it, and five bits unused. Continue in this manner.
The buffer does not have to be one byte. 16-bit or 32-bit buffers are common and are more efficient. You write out bytes whenever the bits therein are eight or more, and shift the remaining 0-7 bits to the start of the buffer.
The only tricky part is what to do at the end, since you may have unused bits in your last byte. Your Huffman codes should have an end symbol to mark the end of the stream. Then you know when you should stop looking for more Huffman codes. If you do not have an end code, then you need to assure somehow that either the remaining bits in the byte cannot be a complete Huffman code, or you need to indicate in some other way where the stream of bits end.