Left Side View of Binary Tree in DSA C# - Time & Space Complexity

Time Complexity: Left Side View of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the left side view of a binary tree changes as the tree grows.

How does the number of nodes affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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;
}
    

This code finds the left side view by visiting each level of the tree and recording the first node at that level.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Visiting each node once in a breadth-first manner using a queue.
• How many times: Each node is enqueued and dequeued exactly once.

How Execution Grows With Input

As the number of nodes increases, the code visits each node once, so the work grows directly with the number of nodes.

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

Pattern observation: The operations grow linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the left side view grows in direct proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "The algorithm only looks at the left children, so it runs faster than visiting all nodes."

[OK] Correct: The code visits every node because it must check both left and right children to know the first node at each level. Skipping right children would miss nodes visible from the left side.

Interview Connect

Understanding this linear time complexity helps you explain how tree traversal works efficiently and shows you can analyze common tree problems clearly.

Self-Check

"What if we changed the traversal to depth-first search instead of breadth-first? How would the time complexity change?"

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