Inorder traversal gives sorted order in Data Structures Theory - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Why does inorder traversal produce sorted order in a BST?

Consider a binary search tree (BST). Why does performing an inorder traversal on it produce the nodes in sorted order?

A Because inorder traversal visits nodes in left-root-right order, which respects BST property of left < root < right.

B Because inorder traversal visits nodes in root-left-right order, which sorts the nodes.

C Because inorder traversal visits nodes randomly, but BST structure sorts them automatically.

D Because inorder traversal visits nodes in right-root-left order, which reverses the sorted order.

💡 Hint

Think about how the BST property relates to the order of visiting nodes.

Inorder traversal visits the left subtree first, then the root, then the right subtree. Since in a BST all left subtree nodes are smaller than the root and all right subtree nodes are larger, this order naturally produces sorted values.

📋 Factual

intermediate

What is the output of inorder traversal on this BST?

Given the BST below, what is the inorder traversal output?

5
/ \
3 7
/ \ \
2 4 8

A [3, 2, 4, 5, 7, 8]

B [8, 7, 5, 4, 3, 2]

C [2, 3, 4, 5, 7, 8]

D [5, 3, 2, 4, 7, 8]

💡 Hint

Remember inorder is left, root, right.

Inorder traversal visits nodes in left-root-right order. For this tree, it visits 2, then 3, then 4, then 5, then 7, then 8.

🔍 Analysis

advanced

What error occurs if inorder traversal is done on a non-BST?

If you perform an inorder traversal on a binary tree that is NOT a binary search tree, what can you say about the output?

A The output will be reversed sorted order.

B The traversal will cause a runtime error because inorder requires BST property.

C The output will always be sorted because inorder traversal visits nodes in order.

D The output may not be sorted because the tree does not guarantee left < root < right.

💡 Hint

Think about what property makes inorder traversal produce sorted output.

Inorder traversal visits nodes in left-root-right order regardless of tree type. Only BSTs guarantee that left subtree nodes are smaller and right subtree nodes are larger, so only then is the output sorted.

❓ Comparison

advanced

Compare inorder and preorder traversal outputs on a BST

Given a BST, how does the output of inorder traversal compare to preorder traversal?

A Inorder traversal outputs sorted nodes; preorder outputs nodes in root-left-right order, not necessarily sorted.

B Both inorder and preorder traversals output nodes in sorted order.

C Preorder traversal outputs sorted nodes; inorder outputs nodes in root-left-right order.

D Neither traversal outputs sorted nodes on a BST.

💡 Hint

Recall the order each traversal visits nodes.

Inorder traversal visits left-root-right, producing sorted output on BSTs. Preorder visits root-left-right, which does not guarantee sorted order.

❓ Reasoning

expert

Why does inorder traversal fail to sort if BST property is violated?

Suppose a binary tree has nodes arranged so that some left child is greater than its parent. Why does inorder traversal fail to produce sorted output in this case?

A Because inorder traversal only works on balanced trees, not unbalanced ones.

B Because inorder traversal depends on the BST property that left children are smaller than the parent to produce sorted output.

C Because inorder traversal visits nodes randomly and does not consider node values.

D Because inorder traversal reverses the order of nodes if BST property is violated.

💡 Hint

Think about what the BST property guarantees about node values.

Inorder traversal visits left subtree, then root, then right subtree. If left child is greater than parent, the order left-root-right does not produce sorted values because the BST property is broken.