ISRO CS 2025 Q.1: solution and answer

Official answer: CSource: ISRO's final answer key of 15 May 2026

1. Inorder traversal visits the left subtree, then the node itself, then the right subtree, applied recursively at every node.

2. Structure from the diagram: $a$ has left child $b$ and right child $c$. $b$ has left child $d$ and right child $e$. $d$ has only a left child $h$. $e$ has left child $i$ and right child $j$. $c$ has left child $f$ and right child $g$. $g$ has only a left child $k$.

3. Left subtree of $a$ (rooted at $b$): its left subtree is rooted at $d$. $d$'s left subtree is just $h$, so the order is $h$, then $d$ (no right child). After that comes $b$ itself.

4. Right subtree of $b$ (rooted at $e$): visit $i$, then $e$, then $j$. So the whole left subtree of $a$ gives $h,d,b,i,e,j$.

5. Then visit the root $a$.

6. Right subtree of $a$ (rooted at $c$): visit $f$ (left), then $c$, then the subtree of $g$. In that subtree $k$ is the left child, so $k$ comes before $g$. This gives $f,c,k,g$.

7. Joining the three parts: $h,d,b,i,e,j,a,f,c,k,g$, which is option (c).

8. Option (a) starts at the root, so it is a preorder-type order. Option (b) ends with the root $a$, so it is a postorder-type order. Option (d) visits $e$ before $i$ and $j$, but in an inorder traversal $e$ must come between its left child $i$ and its right child $j$.