Given a binary tree whose inorder and preorder traversal are given by Inorder…

Concept. In a preorder traversal the first symbol of any block is the root of that subtree. In an inorder traversal that same root splits its block into two parts: every symbol before it belongs to the left subtree and every symbol after it belongs to the right subtree. Applying these two facts together, recursively, reconstructs the tree uniquely. A postorder traversal then reports each subtree as left subtree, right subtree, root.

Application to this item.

1. Preorder begins with B, so B is the root. In the inorder string EICFBGDJHK, B splits it as EICF | B | GDJHK, so the left subtree holds {E, I, C, F} and the right subtree holds {G, D, J, H, K}.

2. The left subtree has 4 nodes, so it takes the next 4 preorder symbols: C, E, I, F. Its root is C. In its inorder part EI | C | F, C has left {E, I} and right {F}, so F is a leaf.

3. For {E, I}: preorder E, I makes E the root, and inorder E | I puts I after E, so E has no left child and I as its right child.

4. The remaining preorder D, G, H, J, K is the right subtree. Its root is D. In its inorder part G | D | JHK, D has left {G} and right {J, H, K}, so G is a leaf.

5. For {J, H, K}: preorder H, J, K makes H the root, and inorder J | H | K gives H the left child J and the right child K.

The reconstructed tree.

Cross-check. Reading this tree in inorder gives E I C F B G D J H K and in preorder gives B C E I F D G H J K, which reproduce both sequences supplied in the question, so the reconstruction is the right one.

Postorder. Visiting left subtree, then right subtree, then root: the C-subtree yields I, E (E after its right child I), then F, then C, i.e. IEFC. The D-subtree yields G, then J, K, H, then D, i.e. GJKHD. Finally the root B. The postorder traversal is therefore IEFCGJKHDB.