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Solution

Given,

Preorder traversal:

$ABCEIFJDGHKL$

Inorder traversal:

$EICFJBGDKHLA$

In preorder traversal, the first element is always the root.

So, root is:

$A$

In inorder traversal, $A$ is at the end.

So, all remaining nodes are in the left subtree of $A$.

Now, after $A$, the next root in preorder is:

$B$

In inorder traversal:

$EICFJBGDKHL$

Here, nodes before $B$ are in left subtree of $B$:

$EICFJ$

Nodes after $B$ are in right subtree of $B$:

$GDKHL$

Now solve left subtree of $B$.

Preorder part is:

$CEIFJ$

Root is $C$.

Inorder part is:

$EICFJ$

Left of $C$ is $EI$ and right of $C$ is $FJ$.

For $EI$, root is $E$, so postorder is:

$IE$

For $FJ$, root is $F$, so postorder is:

$JF$

So, postorder of subtree rooted at $C$ is:

$IEJFC$

Now solve right subtree of $B$.

Preorder part is:

$DGHKL$

Root is $D$.

Inorder part is:

$GDKHL$

Left of $D$ is $G$ and right of $D$ is $KHL$.

So, left postorder is:

$G$

For right part $KHL$, root is $H$.

Left of $H$ is $K$ and right of $H$ is $L$.

So, postorder of subtree rooted at $H$ is:

$KLH$

Thus, postorder of subtree rooted at $D$ is:

$GKLHD$

Now, postorder of subtree rooted at $B$ is:

$IEJFCGKLHDB$

Finally, root $A$ comes at the end.

So, final postorder traversal is:

$IEJFCGKLHDBA$