Question: Show that if the average depth of an n-node binary search tree is Theta(log n) then the height of the tree is O(Squareroot n log

Show that if the average depth of an n-node binary search tree is Theta(log n) then the height of the tree is O(Squareroot n log n). Next demonstrate that this bound is tight: there is a binary search tree with average node depth Theta(log n) and height Theta(Squareroot n log n)

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