Asymptotic complexity of building a binary tree

Question

What is the complexity of building a balanced binary tree of size n from scratch?

Node insertion is O(log n).

However, as you go along, the cumulative time is

O( (log 1) + (log 2) + ... + (log (n-1)) + (log n) ).

What does this "add up" to?

It is o(n log(n)), small 'o', i don't know how far small.

Answer 1

log(1) + log(2) + ... + log(n) = log(1*2*..*n) = log(n!) which is in Theta(nlogn)

So, building a (balanced) tree from scratch has a tight asymptotic bound of Theta(nlogn)

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EDIT:
I also wanted to show you a proof that you cannot build a BST better than O(nlogn) with any algorithm. Basically - this is because it will allow sorting better than O(nlogn) by simply building your tree with the 'better' algorithm, and then - doing in-order traversal on the tree and outputting elements. The result will be a sorted array.
Complexity in-order traversal is O(n), and since assuming the tree is built faster then O(nlogn) - we get contradiction to the fact that sorting is Omega(nlogn) problen.