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

Overview - Left Side View of Binary Tree

What is it?

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

Why it matters

Without the left side view, it is harder to quickly understand the shape and depth of a tree from one side. This concept helps in visualizing and debugging tree structures, and it is useful in problems where side visibility or level order traversal matters. It also helps in applications like graphical tree displays or network routing where perspective matters.

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 related views like right side view, top view, and bottom view of trees, or advanced tree algorithms like vertical order traversal.

Mental Model

Core Idea

The left side view shows the first node visible at each level when looking at the tree 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 you can spot on each floor from your side.

Binary Tree Levels (L) and Left Side View (V):

Level 0:        1          <--- V: 1
               /   \
Level 1:     2       3      <--- V: 2
             / \     \
Level 2:   4   5     6      <--- V: 4

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

Build-Up - 7 Steps

1

Foundation

Understanding Binary Tree Structure

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

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, a root node 1 can have left child 2 and right child 3.

Result

You can visualize a tree as layers of nodes connected by branches.

Understanding the basic shape and connections in a binary tree is essential before exploring 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 one level at a time, from left to right. For example, for the tree: 1 / \ 2 3 You visit 1 first, then 2 and 3. This is often done using a queue to keep track of nodes at each level.

Result

You get a list of nodes grouped by their depth in the tree.

Level order traversal is the foundation for extracting views like the left side view.

3

Intermediate

Extracting Left Side View Using Level Order

🤔Before reading on: do you think the left side view is the first or last node at each level? Commit to your answer.

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

When doing level order traversal, the first node you visit at each level is the leftmost node visible from the left side. By recording only the first node at each level, you get the left side view. Example: For level 0, node 1 is first; for level 1, node 2 is first; for level 2, node 4 is first.

Result

Left side view nodes: [1, 2, 4]

Knowing that the left side view corresponds to the first node at each level simplifies the problem to a simple traversal with a condition.

4

Intermediate

Implementing Left Side View in TypeScript

🤔Before reading on: do you think a recursive or iterative approach is easier for left side view? Commit to your answer.

Concept: Use a queue to perform level order traversal and pick the first node at each level.

TypeScript code example: class TreeNode { val: number; left: TreeNode | null; right: TreeNode | null; constructor(val: number) { this.val = val; this.left = null; this.right = null; } } function leftSideView(root: TreeNode | null): number[] { if (!root) return []; const result: number[] = []; const queue: TreeNode[] = [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); // first node at this level if (node.left) queue.push(node.left); if (node.right) queue.push(node.right); } } return result; } // Example tree: // 1 // / \ // 2 3 // / \ \ // 4 5 6 const root = new TreeNode(1); root.left = new TreeNode(2); root.right = new TreeNode(3); root.left.left = new TreeNode(4); root.left.right = new TreeNode(5); root.right.right = new TreeNode(6); console.log(leftSideView(root));

Result

[1, 2, 4]

Implementing with a queue and checking the first node at each level is a clean, efficient way to get the left side view.

5

Intermediate

Recursive Approach Using Depth Tracking

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

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

In recursion, you visit nodes depth-first. By visiting the left child before the right child, the first node you visit at each depth is the leftmost node. Example TypeScript code: function leftSideViewRecursive(root: TreeNode | null): number[] { const result: number[] = []; function dfs(node: TreeNode | null, depth: number) { if (!node) return; if (depth === result.length) { result.push(node.val); // first node at this depth } dfs(node.left, depth + 1); dfs(node.right, depth + 1); } dfs(root, 0); return result; } console.log(leftSideViewRecursive(root));

Result

[1, 2, 4]

Recursion with depth tracking leverages the call stack to naturally visit nodes in left-to-right order per level.

6

Advanced

Handling Edge Cases and Empty Trees

🤔Before reading on: what should the left side view be for an empty tree? Commit to your answer.

Concept: Consider trees with no nodes or only one side filled to ensure the algorithm handles all cases gracefully.

If the tree is empty (root is null), the left side view is an empty list []. For trees with only left or only right children, the left side view still picks the first node at each level, which may be the only node. Example: Single node tree: 1 Left side view: [1] Right skewed tree: 1 \ 2 \ 3 Left side view: [1, 2, 3]

Result

Empty tree -> [] Single node -> [1] Right skewed -> [1, 2, 3]

Handling edge cases ensures your solution is robust and works for all valid inputs.

7

Expert

Optimizing Space and Time Complexity

🤔Before reading on: do you think left side view can be found in less than O(n) time? Commit to your answer.

Concept: Analyze the algorithm's efficiency and explore if improvements or trade-offs exist.

Both iterative and recursive approaches visit each node once, so time complexity is O(n), where n is the number of nodes. Space complexity depends on the maximum number of nodes stored at once: - Iterative uses a queue that can hold up to the maximum width of the tree. - Recursive uses call stack up to the tree height. No known algorithm can do better than O(n) because every node might be needed to determine the left side view. Optimizations focus on minimizing auxiliary space or combining with other traversals.

Result

Time complexity: O(n) Space complexity: O(w) for iterative (w = max width), O(h) for recursive (h = height)

Understanding complexity helps set realistic expectations and guides optimization efforts.

Under the Hood

The left side view algorithm works by visiting nodes level by level and selecting the first node at each level. Internally, a queue (for iterative) or the call stack (for recursive) keeps track of nodes to visit. The queue stores nodes in order of their appearance in the tree, ensuring the leftmost node is processed first. The recursive approach uses depth to track if a node is the first at its level.

Why designed this way?

This approach was chosen because level order traversal naturally groups nodes by depth, making it easy to pick the first node per level. Alternatives like inorder or preorder traversals do not guarantee level grouping. Using a queue or recursion with depth tracking balances simplicity and efficiency.

Iterative Approach Flow:

[Start]
   |
[Enqueue root]
   |
[While queue not empty]
   |
[For each node in current level]
   |---> [If first node] --> [Add to result]
   |---> [Enqueue left child if exists]
   |---> [Enqueue right child if exists]
   |
[Repeat]
   |
[Return result]

Myth Busters - 3 Common Misconceptions

Quick: Is the left side view the last node at each level? Commit yes or no.

Common Belief: The left side view is the last node you see at each level from the left side.

Reality: The left side view is actually the first node at each level when traversing from left to right.

Why it matters: Choosing the last node would give the right side view, causing incorrect results and confusion.

Quick: Does the left side view include all nodes visible from the left side? Commit yes or no.

Common Belief: The left side view includes all nodes visible from the left side, not just one per level.

Reality: The left side view only includes the first node at each level, not all nodes.

Why it matters: Including all nodes would not be a simple list and would complicate the problem unnecessarily.

Quick: Can you find the left side view without visiting all nodes? Commit yes or no.

Common Belief: You can find the left side view without visiting every node in the tree.

Reality: You must visit all nodes to be sure of the left side view, as any node could be the first at its level.

Why it matters: Skipping nodes risks missing the true leftmost node at some level, leading to wrong output.

Expert Zone

1

The order of visiting left before right child in recursion is critical to ensure the first node at each depth is the leftmost.

2

In very wide trees, the queue size in iterative approach can grow large, impacting memory usage.

3

Combining left side view extraction with other traversals can optimize performance in complex tree problems.

When NOT to use

Left side view is not suitable when you need full visibility of all nodes from the left, or when you want views from other angles like top or bottom. For those, use vertical or horizontal order traversals instead.

Production Patterns

In real systems, left side view helps in graphical tree rendering, debugging tree structures, and in network routing algorithms where visibility from one side matters. It is often combined with caching or incremental updates for dynamic trees.

Connections

Right Side View of Binary Tree

Opposite perspective of the same problem, focusing on the rightmost nodes at each level.

Understanding left side view clarifies how changing traversal order or node selection yields the right side view.

Level Order Traversal

Left side view is a filtered result of level order traversal, selecting only the first node per level.

Mastering level order traversal unlocks many tree view problems including left, right, top, and bottom views.

Human Visual Perception

Both involve seeing the first visible object from a viewpoint, filtering out hidden objects behind.

Knowing how human vision prioritizes front objects helps understand why left side view picks the first node at each level.

Common Pitfalls

#1 Confusing left side view with inorder traversal output.

Wrong approach:

function leftSideView(root) {
if (!root) return [];
const result = [];
function inorder(node) {
if (!node) return;
inorder(node.left);
result.push(node.val);
inorder(node.right);
}
inorder(root);
return result;
}

Correct approach:

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;
}

Root cause:

Misunderstanding that inorder traversal visits nodes in sorted order, not by level or visibility.

#2 Not checking if the tree is empty before processing.

Wrong approach:

function leftSideView(root) {
const result = [];
const queue = [root];
while (queue.length) {
const node = queue.shift();
result.push(node.val);
if (node.left) queue.push(node.left);
if (node.right) queue.push(node.right);
}
return result;
}

Correct approach:

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;
}

Root cause:

Assuming input is always valid and non-null, leading to runtime errors.

Key Takeaways

The left side view of a binary tree is the list of first nodes visible at each level from the left side.

Level order traversal is the key technique to extract the left side view by selecting the first node at each level.

Both iterative (using a queue) and recursive (using depth tracking) methods can find the left side view efficiently.

Handling edge cases like empty or skewed trees ensures your solution is robust.

Understanding the complexity and internal mechanism helps optimize and apply this concept in real-world problems.