Would it be O(26n) where 26 is the number of letters of the alphabet and n is the number of levels of the trie? For example this is the code to print a trie:
public void print()
{
for(int i = 0; i < 26; i++)
{
if(this.next[i] != null)
{
this.next[i].print();
}
}
if(this.word != null)
{
System.out.println(this.word.getWord());
}
}
So watching this code makes me think that my aproximation of the time complexity is correct in the worst of the cases that would be the 26 nodes full for n levels.
Would it be O(26n) where 26 is the number of letters of the alphabet and n is the number of levels of the trie?
No. Each node in the trie must be visited, and O(1) work performed for each one (ignoring the work attributable to processing the children, which is accounted separately). The number of children does not matter on a per-node basis, as long as it is bounded by a constant (e.g. 26).
How many nodes are there altogether? Generally more than the number of words stored in the trie, and possibly a lot more. For a naively-implemented, perfectly balanced, complete trie with n levels below the root, each level has 26 times as many nodes as the previous, and so the total number of nodes is 1 + 26 + 262 + ... + 26n. That is O(26n+1) == O(26n) == O(2n), or "exponential" in the number of levels, which also corresponds to the length of the longest word stored within.
But one is more likely to be interested in measuring the complexity in terms of the number of words stored in the trie. With a careful implementation, it is possible to have nodes only for those words and for each maximal initial substring that is common to two or more of those words. In that event, every node has either zero children or at least two, so for w total words, the total number of nodes is bounded by w + w/2 + w/4 + ..., which converges to 2w. Therefore, a traversal of a trie with those structural properties costs O(2w) == O(w).
Moreover, with a little more thought, it is possible to conclude that the particular structural property I described is not really necessary to have O(w) traversal.