Problem Statement
Application lab in Binary Trees and Traversals
Mission
Count nodes with no left or right child.
Learning outcome: Apply recursion to count leaves
Correctness contract
Invariant: Every non-root node has one parent, and traversal visits every reachable node exactly once.
Required technique: Return one only for a node with no children; otherwise combine counts from both subtrees.
Complexity target: time O(n); space O(height).
Input and output
Input: n followed by n triples: nodeValue leftChildIndex rightChildIndex; -1 means no child. Whitespace may be spaces or line breaks.
Output: Print the single exact numeric result with no label. Return it as a String; Main.java prints it without adding other text.
Assumptions:
- Node 0 is the root.
- Child indexes are -1 or valid indexes, and the encoding is an acyclic binary tree.
Before you code
- Restate the input and output contract, then predict the visible example without running code.
- Implement the core state transition: Return one only for a node with no children; otherwise combine counts from both subtrees.
- Trace the smallest boundary case, verify exact formatting, and justify the authored time and auxiliary-space bounds.
Implement Practice.solve(Scanner sc). Keep every provided filename and public class name unchanged.
Sample input
3 A 1 2 B -1 -1 C -1 -1Sample output
2Why the sample works: Visible walkthrough for the ordinary non-trivial path. Only nodes with neither child contribute one; internal nodes combine the two subtree counts. Input `3 A 1 2 B -1 -1 C -1 -1` therefore produces `2`.
Progressive hints
Try the trace and first milestone before opening a hint. Open them in order.
Open hint 1Hint 1 β Contract: identify what each parsed variable represents and write the invariant beside the loop or recursive method.
Open hint 2Hint 2 β Next step: Trace the smallest non-trivial input and write the structure state after the operation before coding the loop.
Open hint 3Hint 3 β Verification: compare the structure state before and after one operation, then test the smallest valid input and a duplicate or unreachable case when allowed.