Left Side View of Binary Tree in DSA Javascript - Build from Scratch

Left Side View of Binary Tree

📖 Scenario: Imagine you have a family tree represented as a binary tree. You want to see only the leftmost family members at each generation level.

🎯 Goal: Build a program that finds the left side view of a binary tree, showing the first node visible at each level from the left side.

📋 What You'll Learn

Create a binary tree using nodes with val, left, and right properties

Use a variable to keep track of the current level during traversal

Implement a function to collect the leftmost nodes at each level

Print the array of left side view node values

💡 Why This Matters

🌍 Real World

Left side view of a binary tree helps in visualizing hierarchical data from a specific perspective, useful in family trees, organizational charts, and decision trees.

💼 Career

Understanding tree traversal and views is important for software engineers working with hierarchical data structures, UI rendering, and algorithms.

1

Create the Binary Tree

Create a binary tree with a root node called root having value 1. Its left child should have value 2 and right child should have value 3. The left child of node 2 should have value 4, and the right child of node 3 should have value 5.

DSA Javascript

// Create the binary tree nodes and link them
// Your code here

Hint

Use objects with val, left, and right properties to build the tree.

2

Set Up Level Tracking Variable

Create an empty array called leftView to store the left side view node values.

DSA Javascript

const root = {
  val: 1,
  left: {
    val: 2,
    left: { val: 4, left: null, right: null },
    right: null
  },
  right: {
    val: 3,
    left: null,
    right: { val: 5, left: null, right: null }
  }
};

// Create an empty array called leftView
// Your code here

Hint

Use const leftView = []; to create an empty array.

3

Implement Left Side View Function

Write a function called leftSideView that takes node and level as parameters. If node is null, return immediately. If level is equal to the length of leftView, push node.val into leftView. Then recursively call leftSideView for the left child with level + 1, followed by the right child with level + 1. Finally, call leftSideView with root and 0.

DSA Javascript

const root = {
  val: 1,
  left: {
    val: 2,
    left: { val: 4, left: null, right: null },
    right: null
  },
  right: {
    val: 3,
    left: null,
    right: { val: 5, left: null, right: null }
  }
};

const leftView = [];

// Write the leftSideView function and call it with root and 0
// Your code here

Hint

Use recursion to visit nodes level by level, adding the first node at each level to leftView.

4

Print the Left Side View

Print the leftView array using console.log(leftView).

DSA Javascript

const root = {
  val: 1,
  left: {
    val: 2,
    left: { val: 4, left: null, right: null },
    right: null
  },
  right: {
    val: 3,
    left: null,
    right: { val: 5, left: null, right: null }
  }
};

const leftView = [];

function leftSideView(node, level) {
  if (node === null) return;
  if (level === leftView.length) {
    leftView.push(node.val);
  }
  leftSideView(node.left, level + 1);
  leftSideView(node.right, level + 1);
}

leftSideView(root, 0);

// Print the leftView array
// Your code here

Hint

Use console.log(leftView) to see the left side view array.

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