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Its complexity will be O(n^2), not O(n*log(n)).
- Saumesh Kumar March 30, 2014Suppose, you start from root node, then to check that it follows BST property, we have to make sure all of the nodes in its left sub-tree are smaller than root. And all nodes in right sub-tree are greater or equal to it. So effectively we are comparing root's values with n-1 nodes. Therefore, its complexity would be O(n^2).