Left Side View of Binary Tree in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Left Side View for a Simple Binary Tree

What is the output of the following Java code that prints the left side view of a binary tree?

DSA Java

class Node {
    int val;
    Node left, right;
    Node(int val) { this.val = val; }
}

public class Main {
    static void leftSideView(Node root) {
        java.util.Queue<Node> queue = new java.util.LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                Node curr = queue.poll();
                if (i == 0) System.out.print(curr.val + " ");
                if (curr.left != null) queue.add(curr.left);
                if (curr.right != null) queue.add(curr.right);
            }
        }
    }

    public static void main(String[] args) {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.right = new Node(4);
        root.right.right = new Node(5);
        leftSideView(root);
    }
}

A 1 2 5

B 1 2 4

C 1 3 5

D 1 3 4

💡 Hint

The left side view prints the first node at each level from left to right.

The code uses a queue to traverse level by level. It prints the first node at each level, which are 1 (root), 2 (left child of root), and 4 (right child of node 2).

🧠 Conceptual

intermediate

Understanding Left Side View Node Count

Given a binary tree with 4 levels, what is the maximum number of nodes that can appear in its left side view?

A 4

B 8

C 7

D 1

💡 Hint

The left side view shows one node per level.

The left side view shows exactly one node per level, so for 4 levels, the maximum is 4 nodes.

🔧 Debug

advanced

Identify the Bug in Left Side View Code

What error will the following code produce when run on a non-empty binary tree?

DSA Java

void leftSideView(Node root) {
    java.util.Queue<Node> queue = new java.util.LinkedList<>();
    queue.add(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        for (int i = 0; i <= size; i++) {
            Node curr = queue.poll();
            if (i == 0) System.out.print(curr.val + " ");
            if (curr.left != null) queue.add(curr.left);
            if (curr.right != null) queue.add(curr.right);
        }
    }
}

A NullPointerException

B IndexOutOfBoundsException

C No error, prints left side view correctly

D Infinite loop

💡 Hint

Check the for loop condition and how many times poll() is called.

The for loop runs one extra time (i <= size), so poll() returns null on last iteration causing NullPointerException when accessing curr.val.

🚀 Application

advanced

Left Side View of a Skewed Binary Tree

What is the output of the left side view function when run on this skewed binary tree where each node has only a right child: 1 -> 2 -> 3 -> 4?

DSA Java

Node root = new Node(1);
root.right = new Node(2);
root.right.right = new Node(3);
root.right.right.right = new Node(4);
leftSideView(root);

A 1 3 4

B 4

C 1 2 3 4

D 1

💡 Hint

Even if nodes are only on the right, the left side view prints the first node at each level.

Each level has exactly one node, so the left side view prints all nodes: 1 2 3 4.

❓ Predict Output

expert

Output of Left Side View with Null Root

What is the output of the following code when the root node is null?

DSA Java

public static void leftSideView(Node root) {
    if (root == null) {
        System.out.print("Empty");
        return;
    }
    java.util.Queue<Node> queue = new java.util.LinkedList<>();
    queue.add(root);
    while (!queue.isEmpty()) {
        int size = queue.size();
        for (int i = 0; i < size; i++) {
            Node curr = queue.poll();
            if (i == 0) System.out.print(curr.val + " ");
            if (curr.left != null) queue.add(curr.left);
            if (curr.right != null) queue.add(curr.right);
        }
    }
}

public static void main(String[] args) {
    Node root = null;
    leftSideView(root);
}

A No output

B 0

C NullPointerException

D Empty

💡 Hint

Check the null root condition at the start of the function.

The code checks if root is null and prints 'Empty' before returning, so output is 'Empty'.

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 of the tree

Solution

1. Step 1: Understand the view perspective

   The left side view means looking at the tree from the left side, so only nodes visible from that side are included.

2. Step 2: Identify what nodes are visible

   Nodes that are the first in each level from the left are visible, so the leftmost nodes at each depth form the left side view.

3. Final Answer:

   Nodes visible when the tree is viewed from the left side -> Option A

4. Quick Check:

   Left side view = nodes visible from left side [OK]

Hint: Left side view shows first node at each level from left [OK]

Common Mistakes:

• Confusing left side view with right side view
• Including all leaf nodes instead of leftmost nodes
• Thinking it shows deepest nodes only

2. Which Java data structure is best suited to help find the left side view of a binary tree level by level?

easy

A. Queue

B. Stack

C. HashMap

D. ArrayList

Solution

1. Step 1: Identify traversal method for levels

   Level order traversal visits nodes level by level, which is needed to find leftmost nodes at each level.

2. Step 2: Choose data structure for level order

   Queue is used for breadth-first search (level order traversal) because it processes nodes in FIFO order.

3. Final Answer:

   Queue -> Option A

4. Quick Check:

   Level order traversal uses queue [OK]

Hint: Use queue for level order traversal to get left side view [OK]

Common Mistakes:

• Using stack which is for depth-first traversal
• Using HashMap which stores key-value pairs, not order
• Using ArrayList 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, 4, 5]

B. [1, 3, 5]

C. [1, 2, 5]

D. [1, 2, 4]

Solution

1. Step 1: Identify leftmost nodes at each level

   Level 0: Node 1 is leftmost.
   Level 1: Node 2 is leftmost.
   Level 2: Node 4 is leftmost.

2. Step 2: Collect nodes in order

   Left side view nodes are [1, 2, 4].

3. Final Answer:

   [1, 2, 4] -> Option D

4. Quick Check:

   Left side nodes = [1, 2, 4] [OK]

Hint: Pick first node at each level from left [OK]

Common Mistakes:

• Choosing right child nodes instead of left
• Including nodes not visible from left
• Mixing order of nodes

4. Given this Java code snippet for left side view, what is the error?

void leftSideView(Node root) {
  Queue<Node> q = new LinkedList<>();
  q.add(root);
  while (!q.isEmpty()) {
    int size = q.size();
    for (int i = 0; i < size; i++) {
      Node curr = q.poll();
      if (i == 0) System.out.print(curr.data + " ");
      if (curr.left != null) q.add(curr.left);
      if (curr.right != null) q.add(curr.right);
    }
  }
}

medium

A. Printing all nodes instead of only left side nodes

B. Using poll() instead of peek() to get current node

C. No check if root is null before adding to queue

D. Incorrect loop condition in for loop

Solution

1. Step 1: Check root handling

   The code adds root to queue without checking if root is null, which can cause NullPointerException.

2. Step 2: Verify other parts

   Using poll() is correct to remove and get node.
   Printing only when i==0 prints leftmost node.
   Loop condition is correct for level size.

3. Final Answer:

   No check if root is null before adding to queue -> Option C

4. Quick Check:

   Null root check missing causes error [OK]

Hint: Always check if root is null before processing [OK]

Common Mistakes:

• Ignoring null root causing runtime error
• Confusing poll() with peek() usage
• Printing all nodes instead of leftmost only

5. How would you modify the left side view algorithm to return the left side view nodes in reverse order (bottom to top)?

hard

A. Traverse the tree in reverse level order using a queue

B. Use a stack to store leftmost nodes during level order, then pop all

C. Use recursion to collect nodes and reverse the list at the end

D. Print nodes during traversal without storing them

Solution

1. Step 1: Understand normal left side view collection

   Normally, leftmost nodes are collected in order from top to bottom during level order traversal.

2. Step 2: Modify to reverse order

   Store leftmost nodes in a stack during traversal, then pop all to get bottom-to-top order.

3. Final Answer:

   Use a stack to store leftmost nodes during level order, then pop all -> Option B

4. Quick Check:

   Stack reverses order of collected nodes [OK]

Hint: Use stack to reverse order of collected left side nodes [OK]

Common Mistakes:

• Trying to reverse queue order directly
• Not storing nodes before printing
• Using recursion without reversing result