Left Side View of Binary Tree in DSA C#

Mental Model

We want to see the nodes visible when looking at the tree from the left side, which means the first node at each level.

Analogy: Imagine standing on the left side of a tall building with many floors and windows. You only see the first window on each floor from your side, even if there are more windows behind it.

      1
     / \
    2   3
   / \   \
  4   5   6

Left side view: 1 -> 2 -> 4 -> null

Dry Run Walkthrough

Input: Binary tree: 1 -> 2, 3 -> 4, 5, null, 6 (as shown above)

Goal: Find the leftmost node at each level to form the left side view list

Step 1: Start at root node 1, add it to the view

Level 0: [1] -> null

Why: Root is always visible from the left side

Step 2: Go to level 1, nodes are 2 and 3; add leftmost node 2 to view

Level 1: [2] -> 3 -> null

Why: At each level, only the first node from the left is visible

Step 3: Go to level 2, nodes are 4, 5, and 6; add leftmost node 4 to view

Level 2: [4] -> 5 -> 6 -> null

Why: Again, pick the first node at this level

Result:

Left side view: 1 -> 2 -> 4 -> null

Annotated Code

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;
    }
}

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); // first node at this level
                if (node.left != null) queue.Enqueue(node.left);
                if (node.right != null) queue.Enqueue(node.right);
            }
        }
        return result;
    }

    public static void Main() {
        var root = new TreeNode(1,
            new TreeNode(2,
                new TreeNode(4),
                new TreeNode(5)),
            new TreeNode(3,
                null,
                new TreeNode(6)));

        var sol = new Solution();
        var leftView = sol.LeftSideView(root);
        Console.WriteLine(string.Join(" -> ", leftView) + " -> null");
    }
}

if (root == null) return result;

handle empty tree by returning empty list

queue.Enqueue(root);

start BFS from root node

int levelSize = queue.Count;

determine number of nodes at current level

for (int i = 0; i < levelSize; i++) {

iterate over all nodes at this level

var node = queue.Dequeue();

process next node in the current level

if (i == 0) result.Add(node.val);

add first node of level to left side view

if (node.left != null) queue.Enqueue(node.left);

enqueue left child for next level

if (node.right != null) queue.Enqueue(node.right);

enqueue right child for next level

Output

1 -> 2 -> 4 -> null

Complexity Analysis

Time: O(n) because we visit each node exactly once during BFS

Space: O(n) because in worst case the queue holds all nodes at the largest level

vs Alternative: Compared to DFS with tracking levels, BFS naturally processes nodes level by level, making it easier to pick the first node at each level with similar time and space costs

Edge Cases

Empty tree (root is null)

Returns an empty list with no nodes

DSA C#

if (root == null) return result;

Tree with only root node

Returns list with only root value

DSA C#

queue.Enqueue(root);

Tree where some nodes have only right children

Still correctly picks leftmost nodes at each level (which may be right child if no left child)

DSA C#

if (i == 0) result.Add(node.val);

When to Use This Pattern

When asked to find nodes visible from one side of a tree, use level order traversal (BFS) and pick the first node at each level to get the side view.

Common Mistakes

Mistake: Using DFS without tracking level order, causing incorrect node selection for left side view
Fix: Use BFS to process nodes level by level and pick the first node at each level

Mistake: Adding all nodes at a level instead of only the first one
Fix: Add node value only when processing the first node (i == 0) at that level

Summary

Finds the leftmost node at each level of a binary tree to show the left side view.

Use when you need to see which nodes are visible when looking at the tree from the left side.

The key insight is to traverse the tree level by level and pick only the first node at each level.

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