Left Side View of Binary Tree in DSA C++ - 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 from top to bottom. This view helps understand the structure and shape of the tree from a specific angle.

Why it matters

Without the left side view, it is harder to quickly grasp the shape and depth of a tree from a particular perspective. This concept helps in visualizing and debugging tree structures, and it is useful in problems where you need to process or display nodes level by level from one side.

Where it fits

Before learning this, you should understand basic binary trees and tree traversal methods like breadth-first search (BFS) or depth-first search (DFS). After this, you can explore related views like right side view, top view, and bottom view of trees.

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 see on each floor from your position.

Binary Tree Levels:
Level 0:        1
Level 1:      2   3
Level 2:    4   5   6

Left Side View:
1
|
2
|
4

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 also have its own children, forming levels.

Result

You can visualize a tree with nodes connected by branches, starting from the root.

Understanding the basic structure is essential before exploring views or traversals.

2

Foundation

Level Order Traversal Basics

Concept: Learn how to visit nodes level by level using a queue.

Level order traversal visits nodes from top to bottom, left to right at each level. We use a queue to keep track of nodes to visit next.

Result

Nodes are visited in order: root, then all nodes at level 1, then level 2, and so on.

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

3

Intermediate

Extracting Leftmost Node per Level

🤔Before reading on: do you think the left side view always includes the left child of each node? Commit to yes or no.

Concept: Identify the first node visited at each level during level order traversal.

During level order traversal, the first node you visit at each level is the leftmost node visible from the left side. By recording this node, you build the left side view.

Result

You get a list of nodes representing the left side view, e.g., [1, 2, 4].

Knowing that the first node at each level is the left side view node simplifies the problem.

4

Intermediate

Implementing Left Side View Using BFS

🤔Before reading on: do you think adding right children before left children affects the left side view? Commit to yes or no.

Concept: Use a queue to traverse level by level, adding left child before right child to ensure leftmost nodes come first.

Start with the root in the queue. For each level, record the first node. Add left child first, then right child to the queue. Repeat until all levels are processed.

Result

The output is the left side view nodes printed in order.

The order of adding children to the queue controls which node is seen first at each level.

5

Advanced

Implementing Left Side View Using DFS

🤔Before reading on: do you think DFS can find the left side view without level order traversal? Commit to yes or no.

Concept: Use depth-first search with tracking of levels to record the first node visited at each depth.

Traverse the tree starting from root, going left first. Keep track of the current level. If this is the first time visiting this level, record the node. Then visit right child.

Result

You get the left side view nodes by DFS, same as BFS approach.

DFS can also find the left side view by leveraging the order of traversal and level tracking.

6

Expert

Handling Edge Cases and Large Trees

🤔Before reading on: do you think the left side view changes if the tree is skewed or unbalanced? Commit to yes or no.

Concept: Consider trees with missing children, skewed shapes, or very large depth and how the algorithm adapts.

In skewed trees, the left side view is just the nodes along the leftmost path. For large trees, iterative BFS is preferred to avoid stack overflow. Also, handle null nodes carefully to avoid errors.

Result

The left side view correctly reflects the visible nodes regardless of tree shape or size.

Understanding edge cases ensures robust and efficient implementations in real-world scenarios.

Under the Hood

The left side view algorithm works by visiting nodes level by level and selecting the first node encountered at each level. Internally, a queue (for BFS) or recursion stack (for DFS) manages the traversal order. The key is to track levels and record the first node per level, ensuring no duplicates. Memory stores nodes temporarily during traversal, and the output list grows as levels are processed.

Why designed this way?

This approach was chosen because level order traversal naturally processes nodes by depth, making it easy to identify the first node at each level. DFS with level tracking is an alternative that leverages recursion order. Alternatives like full tree scans are less efficient. The design balances simplicity, clarity, and performance.

┌───────────────┐
│    Root (1)   │
└──────┬────────┘
       │
┌──────┴───────┐
│ Level Order  │
│ Traversal Q  │
└──────┬───────┘
       │
┌──────┴─────────────┐
│ 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 contains all left children of nodes.

Reality: The left side view contains the first visible node at each level, which may not always be a left child if the left child is missing.

Why it matters: Assuming all left children appear can cause incorrect outputs when left children are missing or the tree is unbalanced.

Quick: If you add right children before left children in BFS, does the left side view change? Commit to yes or no.

Common Belief: The order of adding children to the queue does not affect the left side view.

Reality: Adding right children before left children can cause the right child to be recorded as the first node at a level, breaking the left side view.

Why it matters: Incorrect child insertion order leads to wrong nodes in the left side view, confusing the tree's shape.

Quick: Can DFS find the left side view without tracking levels? Commit to yes or no.

Common Belief: DFS alone, without level tracking, can produce the left side view.

Reality: DFS must track levels and record the first node visited at each level to correctly produce the left side view.

Why it matters: Without level tracking, DFS may miss nodes or record duplicates, resulting in incorrect views.

Expert Zone

1

The order of child traversal in DFS (left before right) is critical to ensure the first node at each level is the leftmost.

2

In BFS, using a queue size snapshot per level helps isolate nodes per level, preventing mixing nodes from different depths.

3

For very deep trees, iterative BFS avoids stack overflow risks inherent in recursive DFS.

When NOT to use

Avoid using left side view algorithms when you need a full tree representation or other views like top or bottom views. For very large trees where memory is limited, consider streaming or partial traversal methods instead.

Production Patterns

In production, left side view is used in UI tree visualizations, debugging tree structures, and in algorithms that require side-specific node processing, such as certain game AI or hierarchical data rendering.

Connections

Right Side View of Binary Tree

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

Understanding left side view helps grasp right side view by symmetry, reinforcing level order traversal concepts.

Level Order Traversal (Breadth-First Search)

Left side view is a direct application of level order traversal with selective recording.

Mastering level order traversal unlocks many tree view problems, including left side view.

Human Visual Perception

The concept mimics how humans perceive objects from a fixed viewpoint, seeing only the closest visible parts.

Knowing how perspective works in vision helps understand why only certain nodes appear in the left side view.

Common Pitfalls

#1 Adding right child before left child in BFS queue.

Wrong approach:

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

Correct approach:

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

Root cause:

Misunderstanding that child insertion order affects which node is first at each level.

#2 Not tracking levels in DFS, causing duplicate or missing nodes.

Wrong approach:

void dfs(Node* node) {
if (!node) return;
// record node without level check
dfs(node->left);
dfs(node->right);
}

Correct approach:

void dfs(Node* node, int level, vector& view) {
if (!node) return;
if (level == view.size()) view.push_back(node->val);
dfs(node->left, level + 1, view);
dfs(node->right, level + 1, view);
}

Root cause:

Ignoring the need to record only the first node per level.

#3 Assuming left side view includes all left children regardless of tree shape.

Wrong approach:

Always print left child nodes without checking if they are first at their level.

Correct approach:

Print only the first node encountered at each level during traversal.

Root cause:

Confusing left children with visible nodes from the left side.

Key Takeaways

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

Level order traversal (BFS) is the natural method to find the left side view by recording the first node at each level.

Depth-first search (DFS) can also find the left side view by visiting left children first and tracking levels.

The order of child node processing is crucial to correctly identify the leftmost nodes.

Understanding edge cases and traversal order prevents common mistakes and ensures correct results.

Practice

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

easy

A. Nodes at the deepest level of the tree

B. Nodes visible when the tree is viewed from the right side

C. All leaf nodes of the tree

D. Nodes visible when the tree is viewed from the left side

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 encountered 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 D

4. Quick Check:

   Left side view = nodes visible from left [OK]

Hint: Left side view shows first node at each level from left [OK]

Common Mistakes:

• Confusing left side view with right side view
• Thinking it shows all leaf nodes
• Assuming it shows deepest nodes only

2. Which of the following C++ code snippets correctly declares a node structure for a binary tree used in left side view traversal?

easy

A. struct Node { int data; Node* left; Node* right; };

B. struct Node { int data; Node left; Node right; };

C. struct Node { int data; Node* left; int right; };

D. struct Node { int data; int left; int right; };

Solution

1. Step 1: Check pointer usage for child nodes

   Child nodes must be pointers to allow dynamic linking in a tree.

2. Step 2: Validate data types

   Data is int, left and right children are pointers to Node, matching correct binary tree node structure.

3. Final Answer:

   struct Node { int data; Node* left; Node* right; }; -> Option A

4. Quick Check:

   Use pointers for child nodes in tree struct [OK]

Hint: Use pointers for left and right child nodes in struct [OK]

Common Mistakes:

• Using Node instead of Node* for children
• Using int type for child nodes
• Omitting pointers causing infinite size error

3. Consider the following binary tree:

      1
     / \
    2   3
   / \   \
  4   5   6

What is the left side view output of this tree?

medium

A. [1, 3, 6]

B. [1, 2, 4]

C. [1, 2, 5]

D. [1, 2, 6]

Solution

1. Step 1: Identify nodes visible from left at each level

   Level 1: 1, Level 2: 2 (left child), Level 3: 4 (left child of 2)

2. Step 2: Collect nodes in order

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

3. Final Answer:

   [1, 2, 4] -> Option B

4. Quick Check:

   Left side view nodes = [1, 2, 4] [OK]

Hint: Pick first node at each level from left [OK]

Common Mistakes:

• Choosing right child nodes instead of left
• Including nodes not visible from left
• Mixing nodes from different levels

4. The following C++ function is intended to print the left side view of a binary tree. What is the error?

void leftViewUtil(Node* root, int level, int &maxLevel) {
    if (!root) return;
    if (level > maxLevel) {
        cout << root->data << " ";
        maxLevel = level;
    }
    leftViewUtil(root->left, level + 1, maxLevel);
    leftViewUtil(root->right, level + 1, maxLevel);
}

void leftView(Node* root) {
    int maxLevel = 0;
    leftViewUtil(root, 1, maxLevel);
}

medium

A. The order of recursive calls should be right child first, then left child

B. maxLevel should be initialized to -1 instead of 0

C. No error; function correctly prints left side view

D. The function should print after both recursive calls

Solution

1. Step 1: Analyze traversal order and maxLevel usage

   The function visits left child first, then right child, printing first node at each level.

2. Step 2: Check initialization and printing logic

   maxLevel initialized to 0 and level starts at 1, so first level prints correctly; printing before recursion is correct.

3. Final Answer:

   No error; function correctly prints left side view -> Option C

4. Quick Check:

   Correct preorder traversal prints left view [OK]

Hint: Left child first recursion prints left view correctly [OK]

Common Mistakes:

• Changing recursion order breaks left view logic
• Incorrect maxLevel initialization causing missed nodes
• Printing after recursion misses first nodes

5. Given a binary tree where some levels have missing left children but have right children, which approach ensures correct left side view output?

hard

A. Traverse level order and pick first node at each level

B. Traverse preorder and print nodes only if left child exists

C. Traverse inorder and print nodes at odd levels

D. Traverse postorder and print leaf nodes only

Solution

1. Step 1: Understand problem with missing left children

   Left side view must include visible nodes even if left child is missing but right child exists.

2. Step 2: Choose traversal that captures first visible node per level

   Level order traversal visits nodes level by level, so picking first node at each level ensures correct left side view.

3. Final Answer:

   Traverse level order and pick first node at each level -> Option A

4. Quick Check:

   Level order first node per level = left side view [OK]

Hint: Use level order traversal, pick first node at each level [OK]

Common Mistakes:

• Ignoring right child when left child is missing
• Using preorder which may miss visible nodes
• Confusing inorder or postorder with left view logic