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When constructing a binary search tree, we start with the first value as the root and arrange the remaining values based on the tree property. Here's the step-by-step construction: 1. Add \(8\) as root of the tree 2. Add \(5\) as the left child, since \(5 < 8\) 3. Add \(2\) as left child of \(5\), since \(2 < 5\) 4. Add \(3\) as right child of \(2\), since \(3 > 2\) 5. Add \(13\) as the right child of \(8\), since \(13 > 8\) 6. Add \(21\) as the right child of \(13\), since \(21 > 13\) This binary search tree will look like this: ``` 8 / \ 5 13 / \ 2 21 \ 3 ```

To see how the structure changes based on the order of the input, we first need to sort the given set in ascending order: \[2, 3, 5, 8, 13, 21\] Now, construct a binary search tree using the sorted set: 1. Add \(2\) as the root of the tree 2. Add \(3\) as the right child of \(2\), since \(3 > 2\) 3. Add \(5\) as the right child of \(3\), since \(5 > 3\) 4. Add \(8\) as the right child of \(5\), since \(8 > 5\) 5. Add \(13\) as the right child of \(8\), since \(13 > 8\) 6. Add \(21\) as the right child of \(13\), since \(21 > 13\) Thus, the binary search tree will look like this: ``` 2 \ 3 \ 5 \ 8 \ 13 \ 21 ``` This structure is quite different from the first tree, being less balanced. Constructing the binary search tree by randomly adding the values can lead to an unbalanced tree, which might not have the optimal search complexity. If we had constructed an optimally balanced tree, the tree would have shorter height, leading to faster search, insert, and delete operations.