Why Left Side View of Binary Tree in DSA C#?

The Big Idea

What if you could instantly see the 'shape' of a tree from one side without missing a single branch?

The Scenario

Imagine you have a tall tree in your backyard and you want to see which branches you can spot if you stand on the left side of it. You try to write down all the branches you see from bottom to top, but the tree is big and has many branches overlapping each other.

Doing this by just looking and guessing is like trying to find the left side view of a binary tree manually.

The Problem

Manually checking each branch from the left side is slow and confusing because some branches hide others behind them. You might miss some or write down the wrong ones. It's hard to keep track of which branch is the first visible one at each level.

Without a clear method, you waste time and make mistakes.

The Solution

The Left Side View of a Binary Tree concept helps you find exactly which nodes (branches) are visible from the left side, level by level. It uses a simple step-by-step process to check nodes from top to bottom and left to right, picking only the first node you see at each level.

This way, you get a clear list of visible nodes without confusion or missing any.

Before vs After

✗ Before

void PrintLeftView(Node root) {
  // No clear method, just guess and print nodes
  if (root != null) {
    Console.WriteLine(root.Value);
    if (root.Left != null) PrintLeftView(root.Left);
    else if (root.Right != null) PrintLeftView(root.Right);
  }
}

✓ After

void PrintLeftView(Node root) {
  int maxLevel = 0;
  PrintLeftViewUtil(root, 1, ref maxLevel);
}

void PrintLeftViewUtil(Node node, int level, ref int maxLevel) {
  if (node == null) return;
  if (level > maxLevel) {
    Console.WriteLine(node.Value);
    maxLevel = level;
  }
  PrintLeftViewUtil(node.Left, level + 1, ref maxLevel);
  PrintLeftViewUtil(node.Right, level + 1, ref maxLevel);
}

What It Enables

This concept lets you quickly and correctly see the outline of a tree from the left side, which helps in understanding tree shapes and solving many tree problems easily.

Real Life Example

Think of a family tree where you want to see the oldest ancestor visible from the left side at each generation level. Using the Left Side View helps you list those ancestors clearly without confusion.

Key Takeaways

Manual guessing of visible nodes is slow and error-prone.

Left Side View uses a clear method to pick the first visible node at each level.

This helps visualize and solve tree problems efficiently.

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