Could write this code iteratively:

if k < key(n) then n = left(n) else n = right(n) LOOP

- This code is clearly .
- Not necessarily .
- Remember, we are not assured that the tree is balanced

Min(n) ; leftmost leaf of tree rooted at n while left(n) <> NIL n = left(n) return n

Min is ? O(h)

Max(n) ; rightmost leaf of tree rooted at n while right(n) <> NIL n = right(n) return n

Max is ? O(h)