Binary Tree Inorder Traversal

Starting at the root (1), we go to its right child (2) since the left is null. From node 2, we go to its left child (3). Node 3 has no children, so we add 3 to the list. Then we go back to node 2 and add 2. Finally, we go back to the root and add 1. The result is [1, 3, 2], which is incorrect. Let's re-evaluate.

Correct Inorder Traversal:

1. Start at root (1). Does it have a left child? No.
2. Visit the root node (1). Result: [1].
3. Traverse the right subtree, rooted at 2.
4. At node 2, does it have a left child? Yes, 3.
5. At node 3, does it have a left child? No.
6. Visit node 3. Result: [1, 3].
7. At node 3, does it have a right child? No.
8. Return to node 2. Visit node 2. Result: [1, 3, 2].
9. At node 2, does it have a right child? No.
10. Traversal is complete. Final result: [1, 3, 2].

The tree is empty, so the traversal results in an empty list.

Following the left-root-right pattern recursively results in the nodes being visited in their natural sorted order for a Binary Search Tree, which this example happens to be.