The inorder sequence of a binary tree is ABCDEFG, and its postorder sequence is BDCAFGE, so its pre-order sequence is ( ).

The answer to your question is: ABDECFG.

To obtain the pre-order sequence, we need to start with the root node (which is A in this case), then traverse the left subtree, and finally traverse the right subtree.

From the inorder sequence, we know that the left subtree contains nodes B, C, D, and the right subtree contains nodes E, F, G.

From the postorder sequence, we know that the last node is the root node (which is G in this case), and the second-to-last node is the rightmost node in the left subtree (which is F).

Using this information, we can construct the pre-order sequence as follows:

1. Start with the root node A
2. Traverse the left subtree, which has root node B
3. Traverse the left subtree of B, which has root node D
4. Traverse the right subtree of D, which is empty
5. Return to B and traverse its right subtree, which has root node C
6. Return to A and traverse its right subtree, which has root node E
7. Traverse the left subtree of E, which is empty
8. Traverse the right subtree of E, which has root node F
9. Traverse the right subtree of F, which is empty
10. Traverse the right subtree of G, which is empty

Thus, the pre-order sequence is ABDECFG.