Why BST Over Plain Binary Tree in DSA C - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a value into a Binary Search Tree (BST).

DSA C

if (root == NULL) {
    root = (Node*)malloc(sizeof(Node));
    root->data = [1];
    root->left = root->right = NULL;
}

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

Aroot

Bdata

Cvalue

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'root' instead of 'value' to assign data.

Assigning NULL to data field.

We assign the new value to the data field of the new node when root is NULL.

2fill in blank

medium

Complete the code to decide whether to insert into left or right subtree in BST insertion.

DSA C

if (value < root->data) {
    root->left = insert(root->left, [1]);
} else {
    root->right = insert(root->right, value);
}

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

Aroot

BNULL

Cdata

Dvalue

Attempts:

3 left

💡 Hint

Common Mistakes

Passing 'root' instead of 'value' in recursive call.

Passing 'data' which is undefined here.

We recursively insert the same value into the left subtree if it is less than root's data.

3fill in blank

hard

Fix the error in the BST search function to correctly compare values.

DSA C

if (root == NULL || root->data == [1]) {
    return root;
} else if (value < root->data) {
    return search(root->left, value);
} else {
    return search(root->right, value);
}

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

Aroot

Bvalue

Cdata

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing root->data with root instead of value.

Using NULL in comparison.

We compare root's data with the value we are searching for.

4fill in blank

hard

Fill both blanks to create a function that counts nodes in a BST recursively.

DSA C

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

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

Aroot->left

Broot->right

Croot

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'root' instead of child pointers.

Using NULL which causes no recursion.

We count nodes by adding 1 for current node plus counts of left and right subtrees.

5fill in blank

hard

Fill all three blanks to create a function that finds the minimum value in a BST.

DSA C

int findMin(Node* root) {
    if (root == NULL) {
        return -1; // Empty tree
    }
    while (root->[1] != [2]) {
        root = root->[3];
    }
    return root->data;
}

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

Aleft

BNULL

Cright

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' instead of 'left' to find minimum.

Comparing to 'data' instead of NULL.

The minimum value in BST is found by going left until left child is NULL.

Practice

1. Why is a Binary Search Tree (BST) preferred over a plain Binary Tree for searching operations?

easy

A. Because BST does not require pointers to child nodes

B. Because BST uses less memory than a plain Binary Tree

C. Because BST nodes can have more than two children

D. Because BST keeps elements in a sorted order allowing faster search

Solution

1. Step 1: Understand the structure of BST

   A BST stores elements so that left child nodes are smaller and right child nodes are larger than the parent node, maintaining sorted order.

2. Step 2: Compare search efficiency

   This sorted structure allows searching by comparing values and moving left or right, reducing search time to O(log n) on average, unlike a plain Binary Tree which has no order and requires O(n) time.

3. Final Answer:

   Because BST keeps elements in a sorted order allowing faster search -> Option D

4. Quick Check:

   BST sorted order = faster search [OK]

Hint: BST is sorted, plain tree is not, so search is faster [OK]

Common Mistakes:

• Thinking BST uses less memory than plain tree
• Confusing number of children in BST
• Assuming BST nodes don't have child pointers

2. Which of the following is the correct way to declare a node structure for a BST in C?

easy

A. typedef struct Node { int data; struct Node* left; struct Node* right; } Node;

B. typedef struct Node { int data; struct Node left; struct Node right; } Node;

C. typedef struct Node { int data; Node* left; Node* right; } Node;

D. typedef struct Node { int data; int* left; int* right; } Node;

Solution

1. Step 1: Understand C struct pointer usage

   In C, to refer to the same struct type inside itself, pointers must be used with 'struct Node*'.

2. Step 2: Check each option

   typedef struct Node { int data; struct Node* left; struct Node* right; } Node; correctly uses 'struct Node*' for left and right pointers. typedef struct Node { int data; struct Node left; struct Node right; } Node; incorrectly uses struct Node (not pointer), causing infinite size. typedef struct Node { int data; Node* left; Node* right; } Node; uses 'Node*' without 'struct' keyword which is invalid in C without typedef. typedef struct Node { int data; int* left; int* right; } Node; uses int pointers which is incorrect for child nodes.

3. Final Answer:

   typedef struct Node { int data; struct Node* left; struct Node* right; } Node; -> Option A

4. Quick Check:

   Use 'struct Node*' for child pointers in C [OK]

Hint: Use 'struct Node*' for child pointers in C structs [OK]

Common Mistakes:

• Using struct Node instead of pointer causing infinite size
• Omitting 'struct' keyword in C
• Using int* instead of struct pointers

3. What will be the output of the following C code that inserts values into a BST and then performs an inorder traversal?

typedef struct Node {
    int data;
    struct Node* left;
    struct Node* right;
} Node;

Node* insert(Node* root, int val) {
    if (root == NULL) {
        Node* newNode = (Node*)malloc(sizeof(Node));
        newNode->data = val;
        newNode->left = newNode->right = NULL;
        return newNode;
    }
    if (val < root->data) root->left = insert(root->left, val);
    else root->right = insert(root->right, val);
    return root;
}

void inorder(Node* root) {
    if (root == NULL) return;
    inorder(root->left);
    printf("%d ", root->data);
    inorder(root->right);
}

int main() {
    Node* root = NULL;
    root = insert(root, 10);
    root = insert(root, 5);
    root = insert(root, 15);
    inorder(root);
    return 0;
}

medium

A. 5 10 15

B. 10 5 15

C. 15 10 5

D. 10 15 5

Solution

1. Step 1: Trace insertions into BST

   Insert 10 as root. Insert 5 goes to left of 10. Insert 15 goes to right of 10.

2. Step 2: Perform inorder traversal

   Inorder visits left subtree, root, then right subtree. So it prints 5, then 10, then 15.

3. Final Answer:

   5 10 15 -> Option A

4. Quick Check:

   Inorder traversal of BST = sorted order [OK]

Hint: Inorder traversal of BST prints sorted values [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Not inserting nodes correctly
• Printing nodes in wrong order

4. The following code is intended to insert a value into a BST but causes a segmentation fault. What is the error?

Node* insert(Node* root, int val) {
    if (root == NULL) {
        Node newNode;
        newNode.data = val;
        newNode.left = NULL;
        newNode.right = NULL;
        return &newNode;
    }
    if (val < root->data) root->left = insert(root->left, val);
    else root->right = insert(root->right, val);
    return root;
}

medium

A. Not checking if root is NULL before accessing root->data

B. Using 'newNode' instead of 'newNode*' for pointer

C. Returning address of local variable 'newNode' which goes out of scope

D. Missing malloc to allocate memory for new node

Solution

1. Step 1: Analyze node creation in insert function

   The code creates 'Node newNode;' as a local variable on stack inside the function.

2. Step 2: Understand scope and pointer return

   Returning '&newNode' returns address of local variable which is destroyed after function returns, causing invalid memory access and segmentation fault.

3. Final Answer:

   Returning address of local variable 'newNode' which goes out of scope -> Option C

4. Quick Check:

   Local variable address returned = invalid pointer [OK]

Hint: Never return address of local variable; use malloc [OK]

Common Mistakes:

• Ignoring local variable lifetime
• Not using malloc for dynamic allocation
• Confusing pointer and value types

5. You have a plain Binary Tree and a BST both containing the same 7 unique integer values. Which of the following statements about searching for a value is TRUE?

hard

A. Searching in BST is always faster because it is balanced

B. Searching in BST can be faster on average due to ordered structure, but worst-case can be as slow as plain Binary Tree

C. Searching in plain Binary Tree is faster because it has no order constraints

D. Searching speed is the same for both because they have same number of nodes

Solution

1. Step 1: Understand BST search complexity

   BST search is O(log n) on average if balanced, but can degrade to O(n) if skewed.

2. Step 2: Compare with plain Binary Tree search

   Plain Binary Tree has no order, so search is always O(n) as it may need to check all nodes.

3. Step 3: Analyze given statements

   Searching in BST is always faster because it is balanced is false because BST is not always balanced. Searching in BST can be faster on average due to ordered structure, but worst-case can be as slow as plain Binary Tree correctly states average case advantage and worst case equality. Searching in plain Binary Tree is faster because it has no order constraints is false; plain tree is slower. Searching speed is the same for both because they have same number of nodes is false; structure affects search speed.

4. Final Answer:

   Searching in BST can be faster on average due to ordered structure, but worst-case can be as slow as plain Binary Tree -> Option B

5. Quick Check:

   BST average faster, worst-case equal [OK]

Hint: BST average faster, worst-case can be slow [OK]

Common Mistakes:

• Assuming BST is always balanced
• Thinking plain tree search is faster
• Ignoring worst-case BST behavior