Count Total Nodes in Binary Tree in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the total number of nodes in the binary tree.

DSA C++

int countNodes(Node* root) {
    if (root == nullptr) {
        return 0;
    }
    return 1 + countNodes(root->[1]) + countNodes(root->right);
}

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

Aright

Bleft

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' for the left subtree.

Trying to access a non-existent 'parent' pointer.

The function counts nodes by adding 1 for the current node plus the counts of nodes in the left and right subtrees. The left subtree is accessed by root->left.

2fill in blank

medium

Complete the code to correctly check if the tree is empty.

DSA C++

int countNodes(Node* root) {
    if ([1] == nullptr) {
        return 0;
    }
    return 1 + countNodes(root->left) + countNodes(root->right);
}

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

Aroot->left

Broot->right

Croot

Dnullptr

Attempts:

3 left

💡 Hint

Common Mistakes

Checking left or right child instead of root.

Comparing to a variable instead of nullptr.

To check if the tree is empty, we check if the root pointer itself is nullptr.

3fill in blank

hard

Fix the error in the recursive call to count nodes in the right subtree.

DSA C++

int countNodes(Node* root) {
    if (root == nullptr) {
        return 0;
    }
    return 1 + countNodes(root->left) + countNodes(root->[1]);
}

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

Aright

Bparent

Cchild

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' or 'next' which do not exist in the node structure.

Using 'child' which is not a pointer name.

The right subtree is accessed by root->right. Using any other pointer name will cause an error or wrong result.

4fill in blank

hard

Fill both blanks to complete the function that counts nodes using recursion.

DSA C++

int countNodes(Node* root) {
    if (root == [1]) {
        return 0;
    }
    return 1 + countNodes(root->[2]) + countNodes(root->right);
}

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

Anullptr

Broot

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'root' instead of 'nullptr' in base case.

Using 'right' instead of 'left' for left subtree.

The base case checks if root == nullptr to stop recursion. The left subtree is accessed by root->left.

5fill in blank

hard

Fill all three blanks to complete the recursive node counting function correctly.

DSA C++

int countNodes(Node* root) {
    if (root == [1]) {
        return 0;
    }
    return 1 + countNodes(root->[2]) + countNodes(root->[3]);
}

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

Anullptr

Bleft

Cright

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' which does not exist in node structure.

Mixing up left and right pointers.

The base case checks if root == nullptr. The left and right subtrees are accessed by root->left and root->right respectively.