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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the result array for storing the left side view nodes.

DSA Javascript

function leftSideView(root) {
  const result = [1];
  // rest of the code
}

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

A0

B{}

Cnull

D[]

Attempts:

3 left

💡 Hint

Common Mistakes

Using an object {} instead of an array.

Initializing result as null or 0 which cannot hold multiple nodes.

We use an empty array [] to store the nodes visible from the left side.

2fill in blank

medium

Complete the code to check if the current level is visited for the first time.

DSA Javascript

function dfs(node, level) {
  if (!node) return;
  if (level === [1]) {
    result.push(node.val);
  }
  dfs(node.left, level + 1);
  dfs(node.right, level + 1);
}

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

Anode.val

Blevel

Cresult.length

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing level with level itself instead of result length.

Using node.val in the condition which is incorrect.

We compare the current level with result.length to add the first node at each level.

3fill in blank

hard

Fix the error in the DFS traversal order to get the correct left side view.

DSA Javascript

function dfs(node, level) {
  if (!node) return;
  if (level === result.length) {
    result.push(node.val);
  }
  dfs(node.[1], level + 1);
  dfs(node.right, level + 1);
}

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

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Visiting right child before left child causes wrong nodes in left view.

Using invalid property names like parent or child.

We must visit the left child first to capture the leftmost nodes at each level.

4fill in blank

hard

Fill both blanks to complete the iterative approach using a queue for left side view.

DSA Javascript

function leftSideView(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  while (queue.length > 0) {
    const size = queue.length;
    for (let i = 0; i < size; i++) {
      const node = queue.shift();
      if (i === [1]) {
        result.push(node.val);
      }
      if (node.[2]) queue.push(node.[2]);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

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

A0

B1

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using i === 1 instead of i === 0 to pick the first node.

Enqueuing right child before left child changes the order.

We add the first node at each level (i === 0) and enqueue left child first for left side view.

5fill in blank

hard

Fill all three blanks to complete the recursive left side view function with helper.

DSA Javascript

function leftSideView(root) {
  const result = [];
  function helper(node, level) {
    if (!node) return;
    if (level === [1]) {
      result.push(node.[2]);
    }
    helper(node.[3], level + 1);
    helper(node.right, level + 1);
  }
  helper(root, 0);
  return result;
}

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

Aresult.length

Bval

Cleft

Dlevel

Attempts:

3 left

💡 Hint

Common Mistakes

Using level instead of result.length in condition.

Pushing node instead of node.val.

Recursing right child before left child.

We check if level equals result.length, push node.val, and recurse left child first.

Practice

1. What does the "Left Side View" of a binary tree represent?

easy

A. All leaf nodes of the tree

B. Nodes visible when the tree is viewed from the right side

C. Nodes visible when the tree is viewed from the left side

D. Nodes at the deepest level of the tree

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 C

4. Quick Check:

   Left side view = nodes visible from left [OK]

Hint: Think of standing on the left side and seeing nodes [OK]

Common Mistakes:

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

2. Which JavaScript data structure is best suited to help implement the left side view of a binary tree using level order traversal?

easy

A. Queue

B. Stack

C. Set

D. Map

Solution

1. Step 1: Identify traversal method

   Level order traversal visits nodes level by level, left to right.

2. Step 2: Choose data structure for level order

   A queue is used to process nodes in FIFO order, perfect for level order traversal.

3. Final Answer:

   Queue -> Option A

4. Quick Check:

   Level order traversal uses queue [OK]

Hint: Use queue for breadth-first level traversal [OK]

Common Mistakes:

• Using stack which is LIFO, not suitable for level order
• Choosing set or map which don't maintain order
• Confusing queue with stack

3. What is the output of the following code that prints the left side view of the binary tree?

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function leftSideView(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  while (queue.length) {
    const levelSize = queue.length;
    for (let i = 0; i < levelSize; i++) {
      const node = queue.shift();
      if (i === 0) result.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

const root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.right = new Node(6);

console.log(leftSideView(root));

medium

A. [1, 2, 5]

B. [1, 2, 4]

C. [1, 3, 6]

D. [1, 4, 6]

Solution

1. Step 1: Traverse level by level and pick first node

   Level 1: node 1 (first node), Level 2: node 2, Level 3: node 4

2. Step 2: Collect these nodes in result array

   Result array becomes [1, 2, 4]

3. Final Answer:

   [1, 2, 4] -> Option B

4. Quick Check:

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

Hint: Pick first node at each level during BFS [OK]

Common Mistakes:

• Picking right child instead of left
• Including all nodes instead of first per level
• Confusing order of nodes

4. The following code attempts to print the left side view of a binary tree but produces incorrect output. What is the main bug?

function leftSideView(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  while (queue.length) {
    const levelSize = queue.length;
    for (let i = 0; i <= levelSize; i++) {
      const node = queue.shift();
      if (i === 0) result.push(node.val);
      if (node.left) queue.push(node.left);
      if (node.right) queue.push(node.right);
    }
  }
  return result;
}

medium

A. The for loop condition should be i < levelSize, not i <= levelSize

B. The queue should be a stack for correct traversal

C. The check for i === 0 should be i === levelSize - 1

D. The function should return result.length instead of result

Solution

1. Step 1: Analyze the for loop condition

   The loop runs from i = 0 to i <= levelSize, which means one extra iteration causing shift() on empty queue.

2. Step 2: Correct the loop condition

   It should be i < levelSize to process exactly levelSize nodes per level.

3. Final Answer:

   The for loop condition should be i < levelSize, not i <= levelSize -> Option A

4. Quick Check:

   Loop must run exactly levelSize times [OK]

Hint: Use < not <= for loop over fixed size [OK]

Common Mistakes:

• Using <= causes extra iteration and errors
• Confusing queue with stack for BFS
• Wrong index check for first node

5. Given a binary tree where some nodes only have right children, how would the left side view output look for this tree?

    1
     \
      2
       \
        3
         \
          4

hard

A. []

B. [1]

C. [4]

D. [1, 2, 3, 4]

Solution

1. Step 1: Understand left side view with only right children

   Each level has only one node, which is the right child, but also the first node at that level.

2. Step 2: Collect nodes visible from left side

   Since no left children, the right child nodes are visible from left side as they are the only nodes per level.

3. Final Answer:

   [1, 2, 3, 4] -> Option D

4. Quick Check:

   Single node per level = all nodes visible [OK]

Hint: If no left child, right child is visible from left [OK]

Common Mistakes:

• Assuming left side view is empty without left children
• Ignoring right children as visible nodes
• Thinking only left children count