Left Side View of Binary Tree in DSA Javascript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Left Side View of a Binary Tree?

It is the list of nodes visible when the tree is seen from the left side. Only the first node at each level from the left is included.

beginner

Which tree traversal method is commonly used to get the Left Side View of a Binary Tree?

Level order traversal (Breadth-First Search) is commonly used, visiting nodes level by level and picking the first node at each level.

intermediate

In the Left Side View, if a node has no left child but has a right child, which child appears in the view?

The right child will appear in the left side view if there is no left child at that level, because it is the first visible node from the left at that level.

beginner

What data structure helps to keep track of nodes at each level when finding the Left Side View?

A queue is used to perform level order traversal, helping to process nodes level by level.

intermediate

How do you ensure only the first node at each level is added to the Left Side View during traversal?

By adding the first node encountered at each level to the result before processing other nodes at the same level.

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

A Nodes visible from the right side, one per level

B Nodes visible from the left side, one per level

C All leaf nodes

D All nodes in in-order traversal

The left side view shows the first node visible at each level when looking from the left.

Which traversal technique is best suited to find the Left Side View?

A Level order traversal

B Pre-order traversal

C Post-order traversal

D In-order traversal

Level order traversal visits nodes level by level, making it easy to pick the first node at each level.

If a node has no left child but has a right child, will the right child appear in the Left Side View?

A No, only left children appear

B Only if the tree is balanced

C No, it will be skipped

D Yes, if it is the first node at that level

The first node at each level is included regardless of being left or right child.

What data structure is commonly used to implement level order traversal?

A Stack

B Set

C Queue

D Map

A queue helps process nodes in FIFO order, perfect for level order traversal.

How do you identify the first node at each level during traversal?

A By checking if it is the first node dequeued at that level

B By checking if it has a left child

C By checking if it is a leaf node

D By checking if it is the last node at that level

The first node dequeued at each level is the leftmost node visible from the left side.

Explain how to find the Left Side View of a Binary Tree using level order traversal.

Think about visiting nodes one level at a time and picking the first node you see.

Describe what happens in the Left Side View when a node has no left child but has a right child.

Consider visibility from the left side when left child is missing.

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