Path Sum Root to Leaf in Binary Tree in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the current node is a leaf node.

DSA C++

if (root->left == nullptr && root->right == [1]) {
    return sum == root->val;
}

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

A0

BNULL

Cnullptr

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or NULL instead of nullptr.

Checking only one child pointer.

A leaf node has no left or right children, so both should be nullptr.

2fill in blank

medium

Complete the code to recursively check the left subtree for the path sum.

DSA C++

if (root->left != nullptr) {
    if (hasPathSum(root->left, sum - [1])) {
        return true;
    }
}

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

Aroot->val

Bsum

Croot->left->val

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting sum instead of root->val.

Using the child's value instead of current node's value.

We subtract the current node's value from sum before checking the subtree.

3fill in blank

hard

Fix the error in the base case to handle null nodes correctly.

DSA C++

if (root == [1]) {
    return false;
}

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

Anullptr

BNULL

C0

Droot

Attempts:

3 left

💡 Hint

Common Mistakes

Using NULL or 0 instead of nullptr.

Checking root against root itself.

The base case should check if root is nullptr to avoid dereferencing null pointers.

4fill in blank

hard

Fill both blanks to recursively check left and right subtrees for the path sum.

DSA C++

return hasPathSum(root->[1], sum - root->val) || hasPathSum(root->[2], sum - root->val);

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Aleft

Bright

Cval

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Using val or parent instead of left or right.

Not subtracting root->val from sum.

We check both left and right children recursively for the path sum.

5fill in blank

hard

Fill all three blanks to complete the hasPathSum function.

DSA C++

bool hasPathSum(TreeNode* root, int sum) {
    if (root == [1]) return false;
    if (root->left == nullptr && root->right == nullptr) {
        return sum == root->[2];
    }
    return hasPathSum(root->[3], sum - root->val) || hasPathSum(root->right, sum - root->val);
}

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

Anullptr

Bval

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong pointer names or values.

Not handling null root correctly.

Check for null root, compare sum with root->val at leaf, and recurse left and right.