Tree Traversal Inorder Left Root Right in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of inorder traversal?

Given the binary tree below, what is the output of the inorder traversal (Left, Root, Right)?

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

// Tree structure:
//     2
//    / \
//   1   3

Node root = new Node(2);
root.left = new Node(1);
root.right = new Node(3);

void Inorder(Node node) {
    if (node == null) return;
    Inorder(node.left);
    Console.Write(node.val + " ");
    Inorder(node.right);
}

Inorder(root);

A 3 2 1

B 1 2 3

C 2 1 3

D 1 3 2

💡 Hint

Inorder traversal visits left subtree first, then root, then right subtree.

Inorder traversal means visit left child, then root, then right child.

For the tree: 2 with left child 1 and right child 3, the order is 1, 2, 3.

❓ Predict Output

intermediate

Inorder traversal output for unbalanced tree

What is the output of the inorder traversal for this unbalanced tree?

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

// Tree structure:
//     5
//    / 
//   3   
//    \  
//     4

Node root = new Node(5);
root.left = new Node(3);
root.left.right = new Node(4);

void Inorder(Node node) {
    if (node == null) return;
    Inorder(node.left);
    Console.Write(node.val + " ");
    Inorder(node.right);
}

Inorder(root);

A 3 5 4

B 5 3 4

C 4 3 5

D 3 4 5

💡 Hint

Remember to visit left subtree fully before root, then right subtree.

Inorder traversal visits left subtree, then root, then right subtree.

For this tree, left subtree is 3 with right child 4, so order is 3, 4, then root 5.

🔧 Debug

advanced

Identify the error in inorder traversal code

What error will this code produce when performing inorder traversal?

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

void Inorder(Node node) {
    if (node == null) return;
    Inorder(node.right);
    Console.Write(node.val + " ");
    Inorder(node.left);
}

Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);

Inorder(root);

A 3 1 2

B 2 1 3

C 1 3 2

D SyntaxError

💡 Hint

Check the order of recursive calls in the traversal function.

The code visits right subtree first, then root, then left subtree.

This is a reversed inorder traversal, so output is 3 1 2.

🧠 Conceptual

advanced

How many nodes are visited in inorder traversal?

Given a binary tree with 7 nodes, how many nodes does the inorder traversal visit?

A Depends on the tree structure

B 8

C 7

D 6

💡 Hint

Traversal visits every node exactly once.

Inorder traversal visits every node in the tree exactly once, so for 7 nodes, it visits 7 nodes.

🚀 Application

expert

What is the inorder traversal output of this complex tree?

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

DSA C#

class Node {
    public int val;
    public Node left, right;
    public Node(int v) { val = v; left = right = null; }
}

// Tree structure:
//       10
//      /  \
//     5    15
//    / \     \
//   3   7     20
//      /     /  \
//     6     17  25

Node root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);
root.left.left = new Node(3);
root.left.right = new Node(7);
root.left.right.left = new Node(6);
root.right.right = new Node(20);
root.right.right.left = new Node(17);
root.right.right.right = new Node(25);

void Inorder(Node node) {
    if (node == null) return;
    Inorder(node.left);
    Console.Write(node.val + " ");
    Inorder(node.right);
}

Inorder(root);

A 3 5 6 7 10 15 17 20 25

B 3 6 7 5 10 15 17 20 25

C 10 5 3 7 6 15 20 17 25

D 3 5 7 6 10 15 20 17 25

💡 Hint

Follow left subtree fully, then root, then right subtree recursively.

Inorder traversal visits nodes in ascending order for binary search trees.

Visiting left subtree of 10: 3,5,6,7 then root 10, then right subtree 15,17,20,25.

Practice

1. What is the order of nodes visited in an inorder traversal of a binary tree?

easy

A. Left subtree, Root, Right subtree

B. Root, Left subtree, Right subtree

C. Right subtree, Root, Left subtree

D. Root, Right subtree, Left subtree

Solution

1. Step 1: Understand inorder traversal definition

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

2. Step 2: Apply the order to the tree nodes

   This order ensures nodes are visited in ascending order for binary search trees.

3. Final Answer:

   Left subtree, Root, Right subtree -> Option A

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Remember: Inorder = Left, Root, Right [OK]

Common Mistakes:

• Confusing inorder with preorder or postorder
• Visiting root before left subtree
• Visiting right subtree before root

2. Which of the following C# method signatures correctly represents an inorder traversal method for a binary tree node class named Node?

easy

A. void InorderTraversal(Node root)

B. int InorderTraversal(Node root)

C. Node InorderTraversal()

D. bool InorderTraversal(Node root)

Solution

1. Step 1: Identify the purpose of the method

   An inorder traversal method typically visits nodes and prints or processes them without returning a value, so it should be void.

2. Step 2: Check method signature options

   void InorderTraversal(Node root) uses void return type and accepts a Node parameter, which is correct for traversal.

3. Final Answer:

   void InorderTraversal(Node root) -> Option A

4. Quick Check:

   Traversal methods usually return void [OK]

Hint: Traversal methods usually return void [OK]

Common Mistakes:

• Using return types like int or bool unnecessarily
• Missing the root parameter
• Returning Node instead of void

3. Given the following binary tree:

    2
   / \
  1   3

What is the output of the inorder traversal method that prints node values?

medium

A. 1 3 2

B. 2 1 3

C. 3 2 1

D. 1 2 3

Solution

1. Step 1: Apply inorder traversal on the tree

   Visit left subtree (1), then root (2), then right subtree (3).

2. Step 2: Write the visited nodes in order

   The nodes are visited in order: 1, 2, 3.

3. Final Answer:

   1 2 3 -> Option D

4. Quick Check:

   Inorder traversal output = 1 2 3 [OK]

Hint: Inorder prints nodes in ascending order for BST [OK]

Common Mistakes:

• Printing root before left subtree
• Swapping left and right subtree order
• Confusing with preorder or postorder output

4. Consider this C# inorder traversal method:

void InorderTraversal(Node root) {
    if (root == null) return;
    InorderTraversal(root.Right);
    Console.Write(root.Value + " ");
    InorderTraversal(root.Left);
}

What is the error in this code?

medium

A. It causes infinite recursion

B. It misses the base case for null root

C. It visits right subtree before left subtree, reversing inorder order

D. It does not print the root value

Solution

1. Step 1: Analyze the traversal order in the code

   The code visits root.Right first, then root, then root.Left, which is reverse of inorder.

2. Step 2: Identify the correct inorder order

   Inorder should visit left subtree first, then root, then right subtree.

3. Final Answer:

   It visits right subtree before left subtree, reversing inorder order -> Option C

4. Quick Check:

   Inorder = Left, Root, Right [OK]

Hint: Inorder must visit left subtree before right [OK]

Common Mistakes:

• Swapping left and right subtree calls
• Forgetting to check for null root
• Not printing root value

5. You have a binary tree where each node contains an integer. You want to collect all node values in a list using inorder traversal. Which C# code snippet correctly implements this?

hard

A. void Inorder(Node root, List<int> result) {
if (root == null) return;
result.Add(root.Value);
Inorder(root.Left, result);
Inorder(root.Right, result);
}

B. void Inorder(Node root, List<int> result) {
if (root == null) return;
Inorder(root.Left, result);
result.Add(root.Value);
Inorder(root.Right, result);
}

C. void Inorder(Node root, List<int> result) {
if (root == null) return;
Inorder(root.Right, result);
result.Add(root.Value);
Inorder(root.Left, result);
}

D. void Inorder(Node root, List<int> result) {
if (root == null) return;
Inorder(root.Left, result);
Inorder(root.Right, result);
result.Add(root.Value);
}

Solution

1. Step 1: Recall inorder traversal order

   Inorder visits left subtree, then root, then right subtree.

2. Step 2: Check which code matches this order

   void Inorder(Node root, List<int> result) { if (root == null) return; Inorder(root.Left, result); result.Add(root.Value); Inorder(root.Right, result); } visits left subtree, adds root value, then visits right subtree, correctly collecting values in order.

3. Final Answer:

   void Inorder(Node root, List<int> result) { if (root == null) return; Inorder(root.Left, result); result.Add(root.Value); Inorder(root.Right, result); } -> Option B

4. Quick Check:

   Left, Root, Right order with result.Add(root.Value) in middle [OK]

Hint: Add root value between left and right recursive calls [OK]

Common Mistakes:

• Adding root value before left subtree
• Adding root value after right subtree
• Swapping left and right subtree calls