Left Side View of Binary Tree in DSA Javascript - Time & Space Complexity

Time Complexity: Left Side View of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the left side view of a binary tree changes as the tree grows.

How does the number of nodes affect the work done to get the left side view?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function leftSideView(root) {
  if (!root) return [];
  const result = [];
  const queue = [root];
  while (queue.length > 0) {
    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;
}
    

This code finds the left side view by visiting nodes level by level and picking the first node at each level.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs once per level, and the inner for loop visits every node in the tree exactly once.
• How many times: Each node is processed once, so total operations grow with the number of nodes n.

How Execution Grows With Input

As the tree grows, the number of nodes increases, and the code visits each node once.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The work grows directly with the number of nodes, so doubling nodes roughly doubles the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the left side view grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Since we only pick one node per level, the time is O(height) or O(log n)."

[OK] Correct: Even though we pick one node per level for the result, we still visit every node to know which is first at each level, so the time depends on all nodes, not just levels.

Interview Connect

Understanding this helps you explain how tree traversal works and how to analyze algorithms that visit all nodes once, a key skill in many coding challenges.

Self-Check

"What if we used recursion instead of a queue for level order traversal? How would the time complexity change?"

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