Left Side View of Binary Tree in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the TreeNode class with integer value.

DSA Java

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    TreeNode(int [1]) {
        this.val = val;
    }
}

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

Aval

Bvalue

Cnode

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different parameter name than 'val' causes the assignment to not work as intended.

The constructor parameter should be named 'val' to match the assignment 'this.val = val;'.

2fill in blank

medium

Complete the code to start the left side view traversal using a queue.

DSA Java

Queue<TreeNode> queue = new LinkedList<>();
queue.[1](root);

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

AofferFirst

Bpush

Cadd

Denqueue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'push' which is for stacks, not queues.

Using 'enqueue' which is not a Java Queue method.

The 'add' method is used to insert elements into a Java Queue.

3fill in blank

hard

Fix the error in the loop condition to process nodes level by level.

DSA Java

while (!queue.[1]()) {
    int size = queue.size();
    for (int i = 0; i < size; i++) {
        TreeNode node = queue.poll();
        // process node
    }
}

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

AisEmpty

Bempty

ChasNext

DisNotEmpty

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'empty' which does not exist for Queue.

Using 'hasNext' which is for iterators.

The correct method to check if a queue is empty is 'isEmpty()'.

4fill in blank

hard

Fill both blanks to add left and right children to the queue if they exist.

DSA Java

if (node.[1] != null) queue.[2](node.left);
if (node.right != null) queue.add(node.right);

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

Aleft

Bright

Cadd

Doffer

Attempts:

3 left

💡 Hint

Common Mistakes

Checking 'right' instead of 'left' for the first blank.

Using 'add' instead of 'offer' inconsistently.

Check if 'left' child exists, then add it to the queue using 'offer' method.

5fill in blank

hard

Fill all three blanks to add the first node's value of each level to the result list.

DSA Java

List<Integer> result = new ArrayList<>();
while (!queue.isEmpty()) {
    int size = queue.size();
    for (int i = 0; i < size; i++) {
        TreeNode node = queue.poll();
        if (i == [1]) result.[2](node.[3]);
        if (node.left != null) queue.offer(node.left);
        if (node.right != null) queue.offer(node.right);
    }
}

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

A0

Badd

Cval

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using '1' instead of '0' for the first node index.

Using 'set' instead of 'add' for the list.

Accessing 'value' instead of 'val' field.

The first node in each level has index 0; add its 'val' to the result list using 'add' method.

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