The pre-order traversal of a binary search tree is given by…

Key insight: in a BST the preorder sequence gives root, then the entire left subtree, then the entire right subtree. Use that to split the sequence and then produce post-order (left, right, root).

• Root: the first preorder value is 12. Everything less than 12 before the first value greater than 12 belongs to the left subtree: [8,6,2,7,9,10]. The remaining values are the right subtree preorder: [16,15,19,17,20].

• Left subtree (root 8): split its preorder [8,6,2,7,9,10] into left [6,2,7] and right [9,10]. For subtree rooted at 6: left 2 and right 7 give post-order 2,7,6. For subtree rooted at 9 with right child 10, post-order is 10,9. Combine to get left-subtree post-order 2,7,6,10,9,8.

• Right subtree (root 16): preorder [16,15,19,17,20] splits into left [15] and right [19,17,20]. Left gives 15. Right subtree with root 19 has left 17 and right 20, so its post-order is 17,20,19. Combine to get right-subtree post-order 15,17,20,19,16.

• Final post-order: concatenate left-subtree post-order, right-subtree post-order, then root.

Answer: 2, 7, 6, 10, 9, 8, 15, 17, 20, 19, 16, 12