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.

Practice

1. What is the main result of performing an inorder traversal on a binary search tree (BST)?

easy

A. It visits nodes randomly.

B. It visits nodes in descending sorted order.

C. It visits only leaf nodes.

D. It visits nodes in ascending sorted order.

Solution

1. Step 1: Understand inorder traversal definition

   Inorder traversal visits the left child, then the node itself, then the right child.

2. Step 2: Apply inorder traversal to BST property

   Since BST nodes have smaller values on the left and larger on the right, inorder traversal visits nodes in ascending order.

3. Final Answer:

   It visits nodes in ascending sorted order. -> Option D

4. Quick Check:

   Inorder traversal = ascending order [OK]

Hint: Inorder on BST always gives sorted ascending list [OK]

Common Mistakes:

• Confusing inorder with preorder or postorder
• Thinking traversal visits nodes randomly
• Assuming it visits nodes in descending order

2. Which of the following is the correct sequence of steps in an inorder traversal of a binary tree?

easy

A. Visit left child, then node, then right child

B. Visit node, then left child, then right child

C. Visit right child, then node, then left child

D. Visit left child, then right child, then node

Solution

1. Step 1: Recall inorder traversal order

   Inorder traversal means visiting the left subtree first, then the node, then the right subtree.

2. Step 2: Match the correct sequence

   Visit left child, then node, then right child matches this order exactly: left child, node, right child.

3. Final Answer:

   Visit left child, then node, then right child -> Option A

4. Quick Check:

   Inorder = left, node, right [OK]

Hint: Inorder = left subtree, node, right subtree [OK]

Common Mistakes:

• Mixing up preorder and inorder sequences
• Visiting node before left child
• Visiting right child before node

3. Given the BST below, what is the output of an inorder traversal?

      10
     /  \
    5    15
   / \     \
  3   7     20

medium

A. [20, 15, 10, 7, 5, 3]

B. [10, 5, 3, 7, 15, 20]

C. [3, 5, 7, 10, 15, 20]

D. [5, 3, 7, 10, 15, 20]

Solution

1. Step 1: Perform inorder traversal on the BST

   Visit left subtree (3, 5, 7), then root (10), then right subtree (15, 20).

2. Step 2: Write nodes in visited order

   The order is 3, 5, 7, 10, 15, 20.

3. Final Answer:

   [3, 5, 7, 10, 15, 20] -> Option C

4. Quick Check:

   Inorder traversal output = sorted ascending list [OK]

Hint: Inorder traversal of BST = sorted ascending values [OK]

Common Mistakes:

• Listing nodes in preorder or postorder instead
• Confusing left and right subtree order
• Forgetting to include all nodes

4. A student wrote this inorder traversal code but it does not print nodes in sorted order for a BST. What is the mistake?

def inorder(node):
    if node:
        inorder(node.right)
        print(node.value)
        inorder(node.left)

medium

A. The function visits right child before left child, reversing order.

B. The function misses printing the node value.

C. The function does not check if node is None.

D. The function should print before visiting children.

Solution

1. Step 1: Analyze the traversal order in code

   The code visits right child first, then node, then left child, which is reverse of inorder.

2. Step 2: Understand effect on BST traversal

   This reversed order prints nodes in descending order, not ascending sorted order.

3. Final Answer:

   The function visits right child before left child, reversing order. -> Option A

4. Quick Check:

   Inorder must visit left before right [OK]

Hint: Inorder visits left child before right child [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Printing node value in wrong place
• Not checking for null node

5. You have a binary tree that is not a BST. You perform an inorder traversal and get the sequence [4, 2, 5, 1, 6]. Which of the following is true?

hard

A. The tree is a BST because inorder traversal is sorted.

B. The tree is not a BST because inorder traversal is not sorted.

C. Inorder traversal always gives sorted order regardless of tree type.

D. The tree must be balanced to get this sequence.

Solution

1. Step 1: Check if inorder traversal sequence is sorted

   The sequence [4, 2, 5, 1, 6] is not in ascending order.

2. Step 2: Understand BST property and inorder traversal

   In a BST, inorder traversal always produces a sorted ascending sequence. Since this is not sorted, the tree is not a BST.

3. Final Answer:

   The tree is not a BST because inorder traversal is not sorted. -> Option B

4. Quick Check:

   Inorder sorted = BST; not sorted = not BST [OK]

Hint: Inorder sorted? Yes BST; No not BST [OK]

Common Mistakes:

• Assuming inorder always sorts any tree
• Confusing balanced tree with BST
• Ignoring order of traversal output