Tree Traversal Inorder Left Root Right in DSA C# - Time & Space Complexity

Time Complexity: Tree Traversal Inorder Left Root Right

O(n)

Understanding Time Complexity

We want to understand how the time needed to visit all nodes in a tree grows as the tree gets bigger.

Specifically, we look at the inorder traversal that visits nodes in left-root-right order.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public void InorderTraversal(TreeNode node) {
    if (node == null) return;
    InorderTraversal(node.Left);
    Console.WriteLine(node.Value);
    InorderTraversal(node.Right);
}
    

This code visits every node in the tree in left-root-right order.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Exactly once per node in the tree.

How Execution Grows With Input

As the number of nodes grows, the number of visits grows the same way.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete the traversal grows linearly with the number of nodes.

Common Mistake

[X] Wrong: "The traversal takes more time for unbalanced trees because it visits some nodes multiple times."

[OK] Correct: Each node is visited exactly once regardless of tree shape, so unbalanced trees do not increase the total visits.

Interview Connect

Understanding this traversal's time complexity helps you explain how tree operations scale, a key skill in many coding interviews.

Self-Check

"What if we changed the traversal to preorder (root-left-right)? How would the time complexity change?"

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