Trie saves space, but how?

Question

I am confused as to how the Trie implementation saves space & stores data in most compact form!

If you look at the tree below. When you store a character at any nod

Answer 1

You might deduce that it save space is on a ideal machine where every byte is allocated efficiently. However real machines allocate aligned blocks of memory (8 bytes on Java and 16 bytes on some C++) and so it may not save any space.

Java Strings and collections add relatively high amount of over head so the percentage difference can be very small.

Unless your structure is very large the value of your time out weights the memory cost that using the simplest, most standard and easiest to maintain collection is far more important. e.g. your time can very easily be worth 1000x or more the value of the memory you are try to save.

e.g. say you have 10000 names which you can save 16 bytes each by using a trie. (Assuming this can be proven without taking more time) This equates to 16 KB, which at today's prices is worth 0.1 cents. If your time costs your company $30 per hour, the cost of writing one line of tested code might be $1.

If you have think about it a blink of an eye longer to save 16 KB, its unlikely to be worth it for a PC. (mobile devices are a different story but the same argument applies IMHO)

EDIT: You have inspired me to add an update http://vanillajava.blogspot.com/2011/11/ever-decreasing-cost-of-main-memory.html