Maximum Width of Binary Tree in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the function that calculates the maximum width of a binary tree.

DSA C++

int [1](TreeNode* root);

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AmaxWidth

BmaxDepth

CmaxHeight

DmaxSize

Attempts:

3 left

💡 Hint

Common Mistakes

Using maxDepth or maxHeight which relate to tree height, not width.

Using generic names like maxSize that don't specify width.

The function name should be maxWidth as it clearly indicates the purpose of calculating the maximum width of the binary tree.

2fill in blank

medium

Complete the code to initialize the queue for level order traversal.

DSA C++

std::queue<std::pair<TreeNode*, unsigned long long>> [1];

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Astack

Bq

Clist

Dvec

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'stack' which is a different data structure.

Using 'list' or 'vec' which are not queues.

The queue variable is commonly named 'q' for queue, which is used for level order traversal.

3fill in blank

hard

Fix the error in the loop condition to process nodes level by level.

DSA C++

while (![1].empty()) {

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Astack

Bqueue

Cq

Dlist

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'queue' which is the type, not the variable name.

Using 'stack' or 'list' which are not the queue variable.

The queue variable is named 'q', so the loop should check if 'q' is not empty to continue processing.

4fill in blank

hard

Fill both blanks to correctly calculate the width of the current level.

DSA C++

unsigned long long width = last - [1] + 1;
maxWidth = std::max(maxWidth, width[2]);

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Afirst

B+

C-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition when calculating width.

Using wrong variable names or operators.

Width is calculated as last - first + 1. Then maxWidth is updated with the maximum of current maxWidth and width + 0 (width itself).

5fill in blank

hard

Fill all three blanks to correctly push child nodes into the queue with updated indices.

DSA C++

if (node->left) q.push({node->left, [1] * 2});
if (node->right) q.push({node->right, [2] * 2 [3] 1});

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Aindex

C+

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction instead of addition for right child index.

Using wrong variable names.

For left child, index is doubled. For right child, index is doubled and incremented by 1.