Our first thought might be to do an in-order traversal of the BST and return the second-to-last item. This means looking at every node in the BST. That would take time and space, where h is the max height of the tree (which is lg(n) if the tree is balanced, but could be as much as n if not).
We can do better than time and space.
We can do this in one walk from top to bottom of our BST. This means time (again, that's if the tree is balanced, otherwise).
A clean recursive implementation will take space in the call stack, but we can bring our algorithm down to space overall.
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