Left Side View of Binary Tree in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the function that prints the left side view of a binary tree.

DSA C++

void [1](TreeNode* root) {
    if (!root) return;
    // Implementation here
}

Drag options to blanks, or click blank then click option'

AleftSideView

BleftView

CprintLeftView

DprintRightView

Attempts:

3 left

💡 Hint

Common Mistakes

Using a function name that suggests right side view.

Using a generic name that does not indicate left view.

The function is commonly named 'leftView' to indicate it prints the left side view of the tree.

2fill in blank

medium

Complete the code to declare a queue for level order traversal.

DSA C++

std::queue<TreeNode*> [1];

Drag options to blanks, or click blank then click option'

Astack

Bqueue

Cq

Dlist

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'stack' which is a different data structure.

Using 'list' which is not a queue.

The variable 'q' is commonly used as a short name for a queue in traversal algorithms.

3fill in blank

hard

Fix the error in the condition to check if the current node has a left child.

DSA C++

if (current->[1]) {
    q.push(current->left);
}

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cchild

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Checking 'right' instead of 'left'.

Using invalid member names like 'child' or 'node'.

To check if the current node has a left child, use 'left' pointer.

4fill in blank

hard

Fill both blanks to correctly update the level size and iterate through nodes at that level.

DSA C++

int [1] = q.size();
for (int i = 0; i < [2]; i++) {
    TreeNode* current = q.front();
    q.pop();
    // process current
}

Drag options to blanks, or click blank then click option'

AlevelSize

Bsize

Ccount

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables causing mismatch in loop iterations.

Using variable names that are not descriptive.

Using the same variable 'levelSize' to store the size and iterate ensures correct traversal of one level.

5fill in blank

hard

Fill all three blanks to correctly add the first node's value of each level to the result vector.

DSA C++

if (i == [1]) {
    result.[2](current->[3]);
}

Drag options to blanks, or click blank then click option'

A0

Bpush_back

Cval

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong index instead of 0.

Using incorrect vector method like 'add' or 'append'.

Accessing node value with wrong member name.

The first node at each level has index 0; we add its value to result using push_back and access value with 'val'.

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