Left Side View of Binary Tree in DSA Javascript - Deep Dive

Overview - Left Side View of Binary Tree

What is it?

The Left Side View of a Binary Tree is the set of nodes visible when you look at the tree from its left side. It shows the first node you encounter at each level of the tree, starting from the root down to the deepest leaf. This view helps us understand the structure of the tree from a specific angle.

Why it matters

Without the left side view, we might miss how a tree looks from different perspectives, which is important in many applications like graphical displays, tree traversals, and understanding hierarchical data. It helps in visualizing and debugging tree structures by focusing on nodes that are most prominent from the left side.

Where it fits

Before learning this, you should understand what a binary tree is and how tree traversal works (especially level order traversal). After this, you can explore other tree views like right side view, top view, and bottom view, or dive into tree algorithms like depth calculation and balanced trees.

Mental Model

Core Idea

The left side view of a binary tree is the first node visible at each level when looking from the left side.

Think of it like...

Imagine standing on the left side of a tall building with many floors and windows. The left side view is like seeing the first window on each floor that you can spot from where you stand.

Binary Tree Levels and Left Side View:

Level 0:        1
               /   \
Level 1:     2       3
             / \     / \
Level 2:   4   5   6   7

Left Side View: 1 -> 2 -> 4 -> null

Build-Up - 6 Steps

1

Foundation

Understanding Binary Tree Structure

Concept: Learn what a binary tree is and how nodes are connected.

A binary tree is a structure where each node has up to two children: left and right. The top node is called the root. Each child can have its own children, forming levels. For example, the root is level 0, its children are level 1, and so on.

Result

You can visualize a binary tree as a branching structure with nodes connected left and right.

Understanding the basic structure of a binary tree is essential before exploring any specific views or traversals.

2

Foundation

Level Order Traversal Basics

Concept: Learn how to visit nodes level by level from top to bottom.

Level order traversal visits nodes starting from the root, then all nodes at level 1, then level 2, and so on. This is usually done using a queue to keep track of nodes at each level.

Result

You can list nodes in the order they appear level-wise, e.g., 1, 2, 3, 4, 5, 6, 7 for the example tree.

Level order traversal helps us process nodes level by level, which is key to finding the left side view.

3

Intermediate

Identifying Leftmost Nodes at Each Level

🤔Before reading on: do you think the left side view includes all nodes on the left branch or just the first node at each level? Commit to your answer.

Concept: The left side view includes only the first node encountered at each level during level order traversal.

When traversing level by level, the first node you see at each level is the leftmost node. For example, at level 2, even if there are multiple nodes, only the first one from the left is part of the left side view.

Result

The left side view for the example tree is [1, 2, 4].

Knowing that only the first node at each level matters simplifies the problem and guides the traversal approach.

4

Intermediate

Implementing Left Side View Using Queue

🤔Before reading on: do you think we should add right children before left children to get the left side view? Commit to your answer.

Concept: Use a queue for level order traversal, and at each level, record the first node. Add left child before right child to ensure leftmost nodes come first.

Initialize a queue with the root. For each level, note the first node in the queue. Then enqueue its left child first, then right child. Repeat until all levels are processed.

Result

The output array contains the left side view nodes in order.

Adding left children before right ensures the first node at each level is the leftmost, which is crucial for correct output.

5

Advanced

Recursive Approach for Left Side View

🤔Before reading on: do you think recursion can track levels to find leftmost nodes? Commit to your answer.

Concept: Use recursion with a helper function that tracks the current level and records the first node visited at that level.

Start from root at level 0. For each node, if this is the first time visiting this level, record the node's value. Recurse first on the left child, then right child, increasing level by 1 each time.

Result

The recursive function returns the left side view as an array.

Recursion with level tracking provides a clean alternative to queues and highlights the importance of traversal order.

6

Expert

Handling Sparse Trees and Edge Cases

🤔Before reading on: do you think missing left children affect the left side view? Commit to your answer.

Concept: In trees where some left children are missing, the left side view includes the first visible node at each level, which might be a right child.

For example, if a node has no left child but has a right child, that right child becomes the leftmost visible node at that level. Both iterative and recursive methods must handle this by checking children carefully.

Result

The left side view correctly includes nodes even when left children are missing.

Understanding that the left side view depends on visibility, not just left children, prevents bugs in sparse trees.

Under the Hood

The left side view is computed by traversing the tree level by level and selecting the first node encountered at each level. Internally, a queue or recursion stack keeps track of nodes and their levels. The traversal order ensures that the leftmost nodes are processed first, so the first node at each level is recorded. This relies on the breadth-first search (BFS) pattern or depth-first search (DFS) with level tracking.

Why designed this way?

This approach leverages the natural structure of trees and the properties of BFS and DFS to efficiently find visible nodes. Alternatives like scanning all nodes and filtering would be less efficient. The design balances simplicity and performance, making it easy to implement and understand.

┌───────────────┐
│    Root (1)   │
└──────┬────────┘
       │
┌──────┴───────┐
│ Level Order  │
│ Traversal    │
└──────┬───────┘
       │
┌──────┴─────────────┐
│ Record first node   │
│ at each level       │
└──────┬─────────────┘
       │
┌──────┴─────────────┐
│ Left Side View List │
└────────────────────┘

Myth Busters - 3 Common Misconceptions

Quick: Does the left side view always include all left children? Commit to yes or no.

Common Belief: The left side view always includes every left child node in the tree.

Reality: The left side view includes only the first visible node at each level, which might be a right child if the left child is missing.

Why it matters: Assuming all left children are included leads to incorrect views and bugs when trees are sparse or unbalanced.

Quick: Is the left side view the same as pre-order traversal? Commit to yes or no.

Common Belief: The left side view is the same as visiting nodes in pre-order (root-left-right).

Reality: Pre-order visits nodes depth-first, but the left side view depends on the first node at each level, which requires level order or level tracking.

Why it matters: Confusing traversal orders causes wrong outputs and misunderstanding of tree views.

Quick: Can adding right children before left children produce the correct left side view? Commit to yes or no.

Common Belief: Adding right children before left children in traversal still gives the correct left side view.

Reality: Adding right children first can cause the first node at a level to be a right child, missing the true leftmost node.

Why it matters: Traversal order directly affects which nodes are recorded, so wrong order breaks correctness.

Expert Zone

1

The left side view can be computed during a single traversal without storing all nodes, optimizing memory usage.

2

In recursive solutions, tracking the maximum level visited so far is key to avoid overwriting nodes at the same level.

3

Sparse trees require careful checks for null children to avoid missing visible nodes on the right side at a given level.

When NOT to use

Left side view is not suitable when you need full tree traversal or all nodes at each level. For those, use full level order traversal or in-order traversal. Also, for trees with weighted edges or special properties, other views or algorithms might be more relevant.

Production Patterns

In real systems, left side view helps in graphical tree rendering, UI tree navigation, and debugging hierarchical data. It is often combined with right side or top views to provide comprehensive visualizations. Efficient implementations use iterative BFS with queues for performance.

Connections

Breadth-First Search (BFS)

Left side view uses BFS traversal to process nodes level by level.

Understanding BFS helps grasp why the first node at each level is the leftmost visible node.

Depth-First Search (DFS) with Level Tracking

An alternative method to find the left side view uses DFS while tracking levels.

Knowing DFS with level tracking shows how recursion can replace queues for this problem.

Human Visual Perception

The left side view mimics how humans perceive objects from a specific angle, focusing on the first visible parts.

This connection helps appreciate why the concept of 'visibility' matters in data structures and real-world perception.

Common Pitfalls

#1 Adding right child before left child in queue causes wrong left side view.

Wrong approach:

queue.push(node.right);
queue.push(node.left);

Correct approach:

queue.push(node.left);
queue.push(node.right);

Root cause:

Misunderstanding traversal order leads to recording right nodes before left nodes at each level.

#2 Recording every node at each level instead of only the first.

Wrong approach:

for each node in level:
leftSideView.push(node.value);

Correct approach:

for each level:
leftSideView.push(firstNode.value);

Root cause:

Confusing the left side view with full level traversal output.

#3 Ignoring null checks for children causes runtime errors.

Wrong approach:

queue.push(node.left);
queue.push(node.right); // without checking if node.left or node.right is null

Correct approach:

if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);

Root cause:

Not handling missing children leads to pushing undefined nodes and errors.

Key Takeaways

The left side view shows the first visible node at each level of a binary tree when viewed from the left.

Level order traversal with a queue is the most common way to find the left side view by recording the first node at each level.

Traversal order matters: always add left children before right children to ensure correct visibility.

Recursive depth-first search with level tracking is an elegant alternative to iterative methods.

Handling sparse trees correctly requires checking for missing children to avoid missing visible nodes.

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