Left Side View of Binary Tree in DSA C++ - 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 family members visible from the left side, like looking at the tree from the left edge.

🎯 Goal: Build a program that shows the left side view of a binary tree by printing the nodes visible when looking from the left side.

📋 What You'll Learn

Create a binary tree with the exact structure given

Use a variable to track the current level during traversal

Implement a function to find the left side view using preorder traversal

Print the left side view nodes in order

💡 Why This Matters

🌍 Real World

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

💼 Career

Understanding tree traversals and views is important for software engineers working with data structures, algorithms, and system design.

1

Create the Binary Tree Structure

Create a struct called Node with an int data, and two pointers left and right. Then create the exact binary tree with root node 1, left child 2, right child 3, left child of 2 as 4, and right child of 2 as 5.

DSA C++

// Define struct Node
// Create nodes and link them to form the tree
// Your code here

Hint

Define the Node struct with a constructor. Then create nodes and connect them exactly as described.

2

Add a Variable to Track the Current Level

Create an int variable called maxLevel and set it to 0. This will track the deepest level visited so far during traversal.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* 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);

// Create int maxLevel and set it to 0
// Your code here

Hint

This variable helps to know if we have visited a level before.

3

Implement the Left Side View Function

Write a function called leftView that takes Node* root, int level, and a reference to int maxLevel. Use preorder traversal (visit node, then left, then right). If level is greater than maxLevel, print the node's data and update maxLevel. Then recursively call leftView for left and right children with level + 1.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* 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);

int maxLevel = 0;

void leftView(Node* root, int level, int& maxLevel) {
    // Your code here
}

Hint

Use preorder traversal and print the node only if the current level is greater than maxLevel.

4

Print the Left Side View

Call the leftView function with root, 1 as the starting level, and maxLevel. This will print the left side view nodes.

DSA C++

struct Node {
    int data;
    Node* left;
    Node* right;
    Node(int val) : data(val), left(nullptr), right(nullptr) {}
};

Node* 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);

int maxLevel = 0;

void leftView(Node* root, int level, int& maxLevel) {
    if (root == nullptr) {
        return;
    }
    if (level > maxLevel) {
        std::cout data left, level + 1, maxLevel);
    leftView(root->right, level + 1, maxLevel);
}

leftView(root, 1, maxLevel);

Hint

Call leftView(root, 1, maxLevel); inside main() to print the left side view.

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