CareerCup

Can be done by Inorder traversal technique, which will result in O(N) time.

Another Approach:
Let each node in the BST have a field that returns the number of elements in its left and right subtree. Let the left subtree of node T contain only elements smaller than T and the right subtree only elements larger than or equal to T.

Now, suppose we are at node T:

k == num_elements(left subtree of T), then the answer we're looking for is the value in node T
k > num_elements(left subtree of T) then obviously we can ignore the left subtree, because those elements will also be smaller than the kth smallest. So, we reduce the problem to finding the k - num_elements(left subtree of T) smallest element of the right subtree.
k < num_elements(left subtree of T), then the kth smallest is somewhere in the left subtree, so we reduce the problem to finding the kth smallest element in the left subtree.

This is O(log N) on average (assuming a balanced tree)

int kSmallestInBST(NODE *root, int k) {
	int count;
	count = sizeOfBST(root->left) + 1;
	if( k == count)
		return (root->data);
	else if( k < count)
		return kSmallestInBST(k,root->left);
	else
		return kSmallestInBST(k-count,root->right);
}

int sizeOfBST(NODE *root) {
	if(NULL == root)
		return 0;
	else
		return (sizeOfBST(root->left) + 1 
		        + sizeOfBST(root->right));
}

what if we store the size in tree node we can save sizeofBST recursive calls