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
}

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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];

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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);
}

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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
}

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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]);
}

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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'.