Given the following code snippet, which operation is

Solution

The given code performs Inorder Traversal on a binary tree. In this traversal method, the left subtree is visited first, followed by the root node, and finally the right subtree. The recursive calls in the code explicitly follow this order:

1. traverse(root.left) visits the left subtree.
2. System.out.print(root.data) processes the root node.
3. traverse(root.right) visits the right subtree.

For example, if the binary tree is structured as:         1          / \         2   3        / \         4   5  The traversal order is: 4, 2, 5, 1, 3 . Why Inorder Traversal is Important : In binary search trees (BST), an Inorder Traversal retrieves elements in sorted order. This property is essential in many applications, including sorting and range queries. Why Other Options Are Wrong Option A: Preorder Traversal Preorder traversal visits the root node first, then the left subtree, and finally the right subtree. The traversal sequence would be: 1, 2, 4, 5, 3 . This is not the output of the code. Option B: Postorder Traversal Postorder traversal processes the left subtree, followed by the right subtree, and the root node last. The sequence would be: 4, 5, 2, 3, 1 . Again, this does not match the output. Option D: Level Order Traversal Level Order Traversal visits nodes level by level, from left to right. The sequence for the given tree would be: 1, 2, 3, 4, 5 . This traversal method uses a queue and does not match the given code logic. Option E: Depth First Traversal Depth First Search (DFS) is a broader category encompassing Preorder, Inorder, and Postorder traversals. It is not specific enough to describe the given traversal accurately.