Left Side View of Binary Tree in DSA C# - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Left Side View for a Simple Binary Tree

What is the output of the left side view for the given binary tree after running the code?

DSA C#

public class TreeNode {
    public int val;
    public TreeNode left, right;
    public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

public class Solution {
    public IList<int> LeftSideView(TreeNode root) {
        var result = new List<int>();
        if (root == null) return result;
        var queue = new Queue<TreeNode>();
        queue.Enqueue(root);
        while (queue.Count > 0) {
            int levelSize = queue.Count;
            for (int i = 0; i < levelSize; i++) {
                var node = queue.Dequeue();
                if (i == 0) result.Add(node.val);
                if (node.left != null) queue.Enqueue(node.left);
                if (node.right != null) queue.Enqueue(node.right);
            }
        }
        return result;
    }
}

// Tree structure:
//       1
//      / \
//     2   3
//      \   \
//       5   4

var root = new TreeNode(1,
    new TreeNode(2, null, new TreeNode(5)),
    new TreeNode(3, null, new TreeNode(4))
);
var sol = new Solution();
var output = sol.LeftSideView(root);
System.Console.WriteLine(string.Join(", ", output));

A 1, 2, 5

B 1, 5, 4

C 1, 2, 4

D 1, 3, 4

💡 Hint

Remember, the left side view shows the first node visible at each level from the left side.

The code uses a queue to traverse the tree level by level. At each level, it adds the first node's value to the result list. For the given tree, the first nodes at each level are 1 (root), 2 (left child of root), and 5 (right child of 2).

🧠 Conceptual

intermediate

Understanding Left Side View Node Selection

In the left side view of a binary tree, which nodes are included in the output?

A The leftmost node at each level of the tree

B The rightmost node at each level of the tree

C All nodes that have a left child

D All leaf nodes in the tree

💡 Hint

Think about what you see when you look at the tree from the left side.

The left side view includes only the first node visible at each level when looking from the left side, which is the leftmost node at that level.

🔧 Debug

advanced

Identify the Error in Left Side View Implementation

What error will occur when running this code to get the left side view of a binary tree?

DSA C#

public IList<int> LeftSideView(TreeNode root) {
    var result = new List<int>();
    var queue = new Queue<TreeNode>();
    queue.Enqueue(root);
    while (queue.Count > 0) {
        int levelSize = queue.Count;
        for (int i = 0; i < levelSize; i++) {
            var node = queue.Dequeue();
            if (i == 0) result.Add(node.val);
            if (node.left != null) queue.Enqueue(node.left);
            if (node.right != null) queue.Enqueue(node.right);
        }
    }
    return result;
}

A NullReferenceException when root is null

B System.InvalidOperationException due to dequeueing from empty queue

C No error, code runs correctly

D IndexOutOfRangeException in the for loop

💡 Hint

Check the for loop condition and how many times dequeue is called.

The for loop uses 'i <= levelSize' which causes one extra dequeue call than the number of nodes at that level, leading to dequeueing from an empty queue and throwing InvalidOperationException.

❓ Predict Output

advanced

Output of Left Side View for a Skewed Binary Tree

What is the output of the left side view for the following skewed binary tree?

DSA C#

// Tree structure:
// 1
//  \
//   2
//    \
//     3
//      \
//       4

var root = new TreeNode(1,
    null,
    new TreeNode(2,
        null,
        new TreeNode(3,
            null,
            new TreeNode(4)
        )
    )
);
var sol = new Solution();
var output = sol.LeftSideView(root);
System.Console.WriteLine(string.Join(", ", output));

A 1, 3, 4

B 1

C 4

D 1, 2, 3, 4

💡 Hint

Even if nodes are only on the right, the first node at each level is included.

The left side view includes the first node at each level. Here, each level has only one node, so all nodes 1, 2, 3, 4 are included.

🧠 Conceptual

expert

Time Complexity of Left Side View Algorithm

What is the time complexity of the standard breadth-first search approach to find the left side view of a binary tree with n nodes?

A O(n^2), because each level is processed multiple times

B O(log n), because only leftmost nodes are processed

C O(n), because each node is visited once

D O(1), because only the root node is needed

💡 Hint

Consider how many times each node is enqueued and dequeued.

The algorithm visits each node exactly once during the breadth-first traversal, so the time complexity is linear in the number of nodes, O(n).

Practice

1. What does the "Left Side View" of a binary tree represent?

easy

A. Nodes visible when the tree is viewed from the left side

B. Nodes visible when the tree is viewed from the right side

C. All leaf nodes of the tree

D. Nodes at the deepest level only

Solution

1. Step 1: Understand the view perspective

   The left side view shows nodes visible when looking at the tree from the left side.

2. Step 2: Identify visible nodes

   Only the first node at each level from the left is visible in this view.

3. Final Answer:

   Nodes visible when the tree is viewed from the left side -> Option A

4. Quick Check:

   Left side view = nodes seen from left [OK]

Hint: Think of standing on the left side and looking at the tree [OK]

Common Mistakes:

• Confusing left side view with right side view
• Thinking it shows all leaf nodes
• Assuming it shows nodes at the deepest level only

2. Which C# data structure is best suited to help find the left side view of a binary tree level by level?

easy

A. Stack

B. Dictionary

C. Queue

D. List

Solution

1. Step 1: Identify traversal method

   Left side view is found by level order traversal, which processes nodes level by level.

2. Step 2: Choose data structure for level order

   A queue is used to hold nodes of each level in order for processing.

3. Final Answer:

   Queue -> Option C

4. Quick Check:

   Level order traversal uses queue [OK]

Hint: Use queue for level-by-level tree traversal [OK]

Common Mistakes:

• Using stack which is for depth-first traversal
• Choosing dictionary which stores key-value pairs
• Using list without order control

3. What is the output of the left side view for this binary tree?

    1
   / \
  2   3
 /     \
4       5

medium

A. [1, 2, 4]

B. [1, 3, 5]

C. [4, 2, 1]

D. [1, 2, 5]

Solution

1. Step 1: Identify nodes visible from left side

   At level 1: node 1 is visible; level 2: node 2; level 3: node 4.

2. Step 2: List nodes in order of levels

   Left side view nodes are [1, 2, 4].

3. Final Answer:

   [1, 2, 4] -> Option A

4. Quick Check:

   Left side nodes = [1, 2, 4] [OK]

Hint: Pick first node at each level from left [OK]

Common Mistakes:

• Choosing right side nodes instead
• Including nodes not visible from left
• Reversing order of nodes

4. Consider this C# code snippet to get the left side view of a binary tree:

public IList<int> LeftSideView(TreeNode root) {
    var result = new List<int>();
    if (root == null) return result;
    var queue = new Queue<TreeNode>();
    queue.Enqueue(root);
    while (queue.Count > 0) {
        int size = queue.Count;
        for (int i = 0; i < size; i++) {
            var node = queue.Dequeue();
            if (i == 0) result.Add(node.val);
            if (node.left != null) queue.Enqueue(node.left);
            if (node.right != null) queue.Enqueue(node.right);
        }
    }
    return result;
}

What is the bug in this code?

medium

A. It misses the root node in the result

B. It uses a stack instead of a queue

C. It adds the rightmost node instead of leftmost

D. It correctly returns the left side view

Solution

1. Step 1: Analyze the traversal logic

   The code uses a queue for level order traversal and adds the first node at each level to result.

2. Step 2: Verify correctness of node selection

   Adding node when i == 0 ensures leftmost node at each level is added.

3. Final Answer:

   It correctly returns the left side view -> Option D

4. Quick Check:

   First node per level added = left side view [OK]

Hint: Check if first node at each level is added [OK]

Common Mistakes:

• Confusing queue with stack usage
• Adding last node instead of first
• Skipping root node

5. You want to find the left side view of a binary tree but only have access to a recursive preorder traversal method. How can you modify the recursion to get the left side view?

hard

A. Traverse left child before right, but record only leaf nodes

B. Traverse left child before right, and record the first node visited at each depth level

C. Traverse right child before left, and record all nodes at each level

D. Traverse right child before left, and record the last node visited at each depth level

Solution

1. Step 1: Understand preorder traversal order

   Preorder visits root, then left, then right child recursively.

2. Step 2: Record first node at each depth

   By visiting left child first, the first node at each depth is the leftmost node.

3. Step 3: Skip recording if depth already visited

   Only add node value if this is the first visit at that depth.

4. Final Answer:

   Traverse left child before right, and record the first node visited at each depth level -> Option B

5. Quick Check:

   Left-first preorder + first node per depth = left side view [OK]

Hint: Visit left child first and record first node at each depth [OK]

Common Mistakes:

• Visiting right child first loses leftmost nodes
• Recording all nodes instead of first per level
• Recording only leaf nodes misses intermediate nodes