how is std::set (red/black tree) forward iteration implemented?

if i did an inorder traversal of a balanced BST from smallest to largest value, i'd use a DFS which maintains a stack of size lg(n). but if I needed to find the inorder successor of an arbitrary node, it's a worst case lg(n) operation. but if I wanted to iterate in order, id need to find the inorder successor repeatedly for each node yielding O(n*lg(n)). Does std::set use some trick for inorder iteration, or does it really cost O(n*lg(n)), or is the time cost amortized somehow?