Left Side View of Binary Tree in DSA C# - Build from Scratch

Left Side View of Binary Tree

📖 Scenario: Imagine you have a tree of family members, and you want to see only the leftmost person at each level when you look from the left side.

🎯 Goal: Build a program that shows the left side view of a binary tree by printing the values of the leftmost nodes at each level.

📋 What You'll Learn

Create a binary tree with specific nodes

Use a variable to track the current level

Write a function to find the left side view using recursion

Print the left side view nodes in order

💡 Why This Matters

🌍 Real World

Left side view of a binary tree helps in visualizing hierarchical data from a specific angle, useful in UI layouts and organizational charts.

💼 Career

Understanding tree traversals and views is important for software engineers working with data structures, algorithms, and system design.

1

Create the Binary Tree

Create a class called TreeNode with an int val, and two TreeNode references called left and right. Then create a binary tree with root node value 1, left child 2, right child 3, left child of 2 as 4, and right child of 2 as 5.

DSA C#

// Define TreeNode class and create the binary tree
// Your code here

Hint

Define the TreeNode class first, then create the tree nodes exactly as described.

2

Add a Level Tracker Variable

Create an integer variable called maxLevel and set it to 0. This will help track the deepest level visited so far.

DSA C#

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

TreeNode root = new TreeNode(1,
    new TreeNode(2,
        new TreeNode(4),
        new TreeNode(5)
    ),
    new TreeNode(3)
);

int maxLevel = 0;
// Your code here

Hint

Just create an integer variable named maxLevel and set it to zero.

3

Write the Left Side View Function

Write a recursive function called LeftView that takes a TreeNode node, an int level, and a List<int> result. If node is null, return. If level is greater than maxLevel, add node.val to result and update maxLevel to level. Then call LeftView recursively on node.left and node.right with level + 1.

DSA C#

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

TreeNode root = new TreeNode(1,
    new TreeNode(2,
        new TreeNode(4),
        new TreeNode(5)
    ),
    new TreeNode(3)
);

int maxLevel = 0;

// Write the LeftView function below
// Your code here

Hint

Use recursion to visit nodes. Add the node value to the result list only if the current level is deeper than maxLevel.

4

Print the Left Side View

In the Main method, after calling LeftView, print the values in the leftSide list separated by spaces.

DSA C#

using System;
using System.Collections.Generic;

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

class Program
{
    static int maxLevel = 0;

    static void LeftView(TreeNode node, int level, List result)
    {
        if (node == null) return;
        if (level > maxLevel)
        {
            result.Add(node.val);
            maxLevel = level;
        }
        LeftView(node.left, level + 1, result);
        LeftView(node.right, level + 1, result);
    }

    static void Main()
    {
        TreeNode root = new TreeNode(1,
            new TreeNode(2,
                new TreeNode(4),
                new TreeNode(5)
            ),
            new TreeNode(3)
        );

        List leftSide = new List();
        maxLevel = 0;
        LeftView(root, 1, leftSide);

        // Print the left side view values separated by spaces
        // Your code here
    }
}

Hint

Use a foreach loop to print each value in leftSide followed by a space.

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