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++

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

Node* insert(Node* root, int val) {
    if (root == nullptr) {
        root = new Node{val, nullptr, nullptr};
    } else if (val [1] root->data) {
        root->left = insert(root->left, val);
    } else {
        root->right = insert(root->right, val);
    }
    return root;
}

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

A<

B>

C==

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' instead of '<' causes wrong placement in the tree.

Using '==' leads to incorrect logic for insertion.

In a BST, values less than the current node go to the left subtree.

2fill in blank

medium

Complete the code to search for a value in a BST.

DSA C++

bool search(Node* root, int val) {
    if (root == nullptr) return false;
    if (root->data == val) return true;
    if (val [1] root->data) {
        return search(root->left, val);
    } else {
        return search(root->right, val);
    }
}

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

A>

B>=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>' causes search in wrong subtree.

Using '==' in condition leads to redundant checks.

To search in BST, go left if val is less than current node's data.

3fill in blank

hard

Fix the error in the code that finds the minimum value node in a BST.

DSA C++

Node* findMin(Node* root) {
    while (root->[1] != nullptr) {
        root = root->left;
    }
    return root;
}

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'right' pointer leads to maximum value instead of minimum.

Using 'parent' or 'child' causes compilation errors.

The minimum value in BST is found by going left until no more left child.

4fill in blank

hard

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

DSA C++

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

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

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' or 'child' pointers which do not exist.

Counting only one subtree leads to incorrect total.

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 creates a map of node values to their depths in a BST.

DSA C++

void mapDepths(Node* root, int depth, std::map<int, int>& depthMap) {
    if (root == nullptr) return;
    depthMap[[1]] = [2];
    mapDepths(root->[3], depth + 1, depthMap);
    mapDepths(root->right, depth + 1, depthMap);
}

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

Aroot->data

Bdepth

Cleft

Dright

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong keys or values in the map.

Not recursing on both left and right children.

Map node's data to current depth, then recurse left and right subtrees.

Practice

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

easy

A. Because BST keeps elements in sorted order allowing faster search

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

C. Because BST nodes can have more than two children

D. Because BST does not require pointers to child nodes

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) search.

3. Final Answer:

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

4. Quick Check:

   BST sorted order = faster search [OK]

Hint: BST sorts nodes for quick search, plain tree does not [OK]

Common Mistakes:

• Thinking BST uses less memory than plain tree
• Confusing number of children allowed 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. struct Node { int data; Node* parent; };

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

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

D. struct Node { int data; Node* next; };

Solution

1. Step 1: Identify correct pointer usage for children

   A BST node must have pointers to left and right child nodes, so Node* left and Node* right are needed.

2. Step 2: Check each option

   struct Node { int data; Node* left; Node* right; }; correctly declares pointers to left and right children. struct Node { int data; Node left; Node right; }; incorrectly uses Node left and right (not pointers). struct Node { int data; Node* parent; }; only has parent pointer, missing children. struct Node { int data; Node* next; }; uses next pointer which is for linked lists, not BST.

3. Final Answer:

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

4. Quick Check:

   BST node needs left and right pointers [OK]

Hint: BST nodes have left and right child pointers [OK]

Common Mistakes:

• Using objects instead of pointers for children
• Confusing parent pointer with child pointers
• Using 'next' pointer like in linked lists

3. Consider the following code inserting values into a BST:

#include <iostream>

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

Node* insert(Node* root, int val) {
  if (!root) {
    Node* n = new Node{val, nullptr, nullptr};
    return n;
  }
  if (val < root->data) root->left = insert(root->left, val);
  else root->right = insert(root->right, val);
  return root;
}

void inorder(Node* node) {
  if (!node) return;
  inorder(node->left);
  std::cout << node->data << " ";
  inorder(node->right);
}

int main() {
  Node* root = nullptr;
  root = insert(root, 10);
  root = insert(root, 5);
  root = insert(root, 15);
  root = insert(root, 7);
  inorder(root);
  return 0;
}

What is the output of the inorder traversal?

medium

A. 10 5 7 15

B. 15 10 7 5

C. 5 7 10 15

D. 7 5 10 15

Solution

1. Step 1: Understand insertion order and BST property

   Values inserted are 10 (root), 5 (left of 10), 15 (right of 10), 7 (right of 5). BST property keeps left < parent < right.

2. Step 2: Perform inorder traversal

   Inorder traversal visits left subtree, node, then right subtree. So nodes visited in order: 5, 7, 10, 15.

3. Final Answer:

   5 7 10 15 -> Option C

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
• Misplacing 7 as left child of 10
• Ignoring BST insertion rules

4. The following code attempts to insert a value into a BST but has a bug:

Node* insert(Node* root, int val) {
  if (root == nullptr) {
    Node* n = new Node{val, nullptr, nullptr};
    return n;
  }
  if (val <= root->data) {
    insert(root->left, val);
  } else {
    insert(root->right, val);
  }
  return root;
}

What is the bug and how to fix it?

medium

A. The recursive calls do not assign returned nodes to root's children; fix by root->left = insert(root->left, val) and root->right = insert(root->right, val)

B. The base case is wrong; should check if val == root->data

C. The function should return void instead of Node*

D. The comparison should be val < root->data instead of val <= root->data

Solution

1. Step 1: Identify missing assignment in recursion

   The recursive calls insert(root->left, val) and insert(root->right, val) do not update root's left or right pointers with the returned new nodes.

2. Step 2: Fix by assigning returned nodes

   Assign the recursive call results back: root->left = insert(root->left, val); root->right = insert(root->right, val); so the tree links update correctly.

3. Final Answer:

   The recursive calls do not assign returned nodes to root's children; fix by root->left = insert(root->left, val) and root->right = insert(root->right, val) -> Option A

4. Quick Check:

   Assign recursive insert results to children [OK]

Hint: Always assign recursive insert results to child pointers [OK]

Common Mistakes:

• Ignoring returned node from recursion
• Changing base case unnecessarily
• Changing comparison operator without reason

5. You have a plain Binary Tree and a BST both containing the same 7 unique values. You want to find if a value exists in the tree. Which of the following statements best explains why searching in the BST is generally faster than in the plain Binary Tree?

hard

A. BST uses hashing internally to find values instantly

B. BST stores values in a linked list internally, making search O(1)

C. Plain Binary Tree nodes have no pointers, so search is impossible

D. BST allows skipping half of the nodes at each step due to ordering, reducing search time to O(log n) on average

Solution

1. Step 1: Understand search in BST vs plain Binary Tree

   BST maintains sorted order so at each node, you decide to go left or right, effectively halving the search space each step.

2. Step 2: Compare time complexities

   This halving leads to average search time O(log n). Plain Binary Tree has no order, so search may check every node, O(n) time.

3. Final Answer:

   BST allows skipping half of the nodes at each step due to ordering, reducing search time to O(log n) on average -> Option D

4. Quick Check:

   BST halves search space each step [OK]

Hint: BST halves search space; plain tree searches all nodes [OK]

Common Mistakes:

• Thinking BST uses linked lists internally
• Believing plain Binary Tree has no pointers
• Assuming BST uses hashing