Left Side View of Binary Tree in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the TreeNode class with left and right children.

DSA C#

public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int x) { val = x; left = [1]; right = null; }
}

Drag options to blanks, or click blank then click option'

Anew TreeNode(0)

Bthis

Cnull

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning left to 'this' which causes self-reference.

Assigning left to a new node with value 0 which is incorrect initialization.

When initializing a new node, its left and right children should start as null because they don't point to any nodes yet.

2fill in blank

medium

Complete the code to add the root node's value to the result list when visiting a new level.

DSA C#

if (level == result.Count) {
    result.[1](node.val);
}

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ARemove

BAdd

CClear

DInsert

Attempts:

3 left

💡 Hint

Common Mistakes

Using Remove which deletes elements instead of adding.

Using Clear which empties the list.

We use Add to append the node's value to the result list when we first visit a new level.

3fill in blank

hard

Fix the error in the recursive call to traverse the left subtree first.

DSA C#

if (node.left != null) {
    LeftView(node.[1], level + 1, result);
}

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Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.right which goes to the right subtree first.

Using node.val which is an integer, not a node.

To get the left side view, we must traverse the left child first, so we use node.left.

4fill in blank

hard

Fill both blanks to complete the recursive calls for left and right children in order.

DSA C#

if (node.[1] != null) {
    LeftView(node.[1], level + 1, result);
}
if (node.[2] != null) {
    LeftView(node.[2], level + 1, result);
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping left and right children order.

Using node.val which is not a child node.

We first traverse the left child, then the right child to capture the left side view correctly.

5fill in blank

hard

Fill all three blanks to complete the method signature and initial call for left side view.

DSA C#

public IList<int> LeftSideView([1] root) {
    var result = new List<int>();
    if (root != null) {
        LeftView(root, [2], result);
    }
    return result;
}

private void LeftView(TreeNode node, int [3], IList<int> result) {
    // recursive logic here
}

Drag options to blanks, or click blank then click option'

ATreeNode

B0

Clevel

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong parameter types or names.

Starting level at 1 instead of 0.

The method takes a TreeNode as root, starts level at 0, and uses level as the parameter name in recursion.

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