Consider the traversal of a tree Preorder → ABCEIFJDGHKL Inorder →…

Postorder traversal: IEJFCGKLHDBA

1. Start with preorder to get the root: preorder begins with A, so A is the root. In the inorder sequence, all nodes left of A form the left subtree.

2. For the left subtree, the next preorder root is B. Inorder splits around B into left part (E I C F J) and right part (G D K H L).

3. Build B's left subtree: root C with inorder left E I and right F J. E has a right child I, so its postorder is I E. F has a right child J, so its postorder is J F. Combining gives the C-subtree postorder: I E J F C.

4. Build B's right subtree: root D with left G and right H (where H has left K and right L). The postorder of D-subtree is G K L H D.

5. Combine for B: left-subtree postorder + right-subtree postorder + B = IEJFC + GKLHD + B = IEJFCGKLHDB. Finally append the root A to get the full postorder: IEJFCGKLHDBA.