BST Property and Why It Matters in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a value into a BST correctly.

DSA C++

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

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

Avalue

Broot->val

Croot

Dnullptr

Attempts:

3 left

💡 Hint

Common Mistakes

Using root->val instead of value causes infinite recursion.

Using nullptr will cause errors because insert expects a value.

We must insert the same value into the left subtree if it is less than the current node's value.

2fill in blank

medium

Complete the condition to check if a BST node's left child is valid.

DSA C++

if (root->left && root->left->val [1] root->val) {
    // valid left child
}

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

A>=

B>

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using > instead of < reverses the BST property.

Using == allows duplicates on the left, which is usually not allowed.

In a BST, the left child's value must be less than the current node's value.

3fill in blank

hard

Fix the error in the BST search function to return true when the value matches the current node.

DSA C++

if (root == nullptr) return false;
if (root->val [1] value) return true;
if (value < root->val) return search(root->left, value);
else return search(root->right, value);

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

A==

B!=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using != causes the function to return true incorrectly.

Using < or > here breaks the logic of finding the exact value.

We return true when the current node's value equals the search value.

4fill in blank

hard

Fill both blanks to correctly check BST property for right child.

DSA C++

if (root->right && root->right->val [1] root->val) {
    return false; // violates BST property
}
return root->left == nullptr || root->left->val [2] root->val;

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= or >= allows duplicates which may violate strict BST rules.

Mixing up the comparison operators for left and right children.

Right child's value must be greater than the current node's value; left child's value must be less.

5fill in blank

hard

Fill all three blanks to create a correct BST validation condition.

DSA C++

bool isValidBST(TreeNode* root) {
    if (!root) return true;
    if (root->left && root->left->val [1] root->val) return false;
    if (root->right && root->right->val [2] root->val) return false;
    return isValidBST(root->left) && isValidBST(root->right) && root->val [3] INT_MAX;
}

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

A<

B>

C<=

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= or >= incorrectly for child comparisons.

Not checking the root value against INT_MAX properly.

Left child must be less than root, right child greater than root, and root value must be less than or equal to INT_MAX.

Practice

1. What is the main property that defines a Binary Search Tree (BST)?

easy

A. For each node, all values in the left subtree are smaller, and all values in the right subtree are larger.

B. Each node has exactly two children.

C. All nodes have the same value.

D. Nodes are arranged in a complete binary tree structure.

Solution

1. Step 1: Understand BST definition

   A BST requires that for every node, values in the left subtree are smaller and values in the right subtree are larger.

2. Step 2: Compare options with BST property

   For each node, all values in the left subtree are smaller, and all values in the right subtree are larger. matches the BST property exactly, while others describe different tree types or incorrect properties.

3. Final Answer:

   For each node, all values in the left subtree are smaller, and all values in the right subtree are larger. -> Option A

4. Quick Check:

   BST property = For each node, all values in the left subtree are smaller, and all values in the right subtree are larger. [OK]

Hint: Remember: left < node < right in BST [OK]

Common Mistakes:

• Thinking BST means every node has two children
• Confusing BST with balanced or complete trees
• Assuming all nodes have the same value

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

easy

A. class Node { int data; Node left; Node right; };

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

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

D. class Node { int data; Node* left; Node* right; };

Solution

1. Step 1: Check pointer usage in node declaration

   BST nodes must use pointers for left and right children to allow dynamic linking.

2. Step 2: Compare struct and class syntax

   struct Node { int data; Node* left; Node* right; }; correctly uses struct with pointers. class Node { int data; Node* left; Node* right; }; uses class but misses access specifiers (defaults private). Declarations using objects instead of pointers for children are incorrect.

3. Final Answer:

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

4. Quick Check:

   Use pointers for children in BST node [OK]

Hint: Use pointers for child nodes in BST struct [OK]

Common Mistakes:

• Using objects instead of pointers for children
• Forgetting semicolon after struct
• Using class without public access for members

3. Consider the following BST insertion code snippet in C++:

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

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

int main() {
  Node* root = nullptr;
  root = insert(root, 10);
  root = insert(root, 5);
  root = insert(root, 15);
  root = insert(root, 12);
  root = insert(root, 18);
  return 0;
}

What is the in-order traversal output of the BST after these insertions?

medium

A. [10, 12, 5, 15, 18]

B. [10, 5, 15, 12, 18]

C. [18, 15, 12, 10, 5]

D. [5, 10, 12, 15, 18]

Solution

1. Step 1: Understand insertion order and BST structure

   Inserted values: 10 (root), 5 (left), 15 (right), 12 (left of 15), 18 (right of 15).

2. Step 2: Perform in-order traversal (left, root, right)

   Traverse left subtree: 5, root: 10, left of right subtree: 12, root of right subtree: 15, right of right subtree: 18. So output is [5, 10, 12, 15, 18].

3. Final Answer:

   [5, 10, 12, 15, 18] -> Option D

4. Quick Check:

   In-order traversal of BST = sorted values [OK]

Hint: In-order traversal of BST gives sorted values [OK]

Common Mistakes:

• Confusing pre-order or post-order with in-order
• Ignoring BST insertion rules
• Mixing up left and right subtree traversal

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

What is the main issue with this code regarding BST property?

medium

A. It does not allocate memory for new nodes.

B. Using <= allows duplicate values on the left, which may violate strict BST property.

C. It uses recursion incorrectly causing infinite loops.

D. It does not handle the case when root is null.

Solution

1. Step 1: Analyze the condition for left subtree insertion

   The code uses val <= root->data to decide left subtree insertion, allowing duplicates on the left.

2. Step 2: Understand BST strict property

   Strict BST property requires all left subtree values to be strictly less than the node's value. Allowing duplicates on left can break this.

3. Final Answer:

   Using <= allows duplicate values on the left, which may violate strict BST property. -> Option B

4. Quick Check:

   Strict BST: left < node < right [OK]

Hint: Use < for left subtree to avoid duplicates breaking BST [OK]

Common Mistakes:

• Ignoring duplicate handling in BST
• Thinking recursion causes infinite loops here
• Assuming null root case is missing

5. You have a BST where each node stores an integer. You want to find the kth smallest element efficiently. Which approach best uses the BST property to solve this?

hard

A. Perform an in-order traversal and stop when the kth element is reached.

B. Perform a level-order traversal and count nodes until kth is found.

C. Traverse only the right subtree k times to find the kth smallest.

D. Use pre-order traversal and store all nodes in a list, then sort it.

Solution

1. Step 1: Recall BST in-order traversal property

   In-order traversal visits nodes in ascending order for BST.

2. Step 2: Use in-order traversal to find kth smallest

   By traversing in-order and counting nodes, we can stop at kth node, efficiently finding kth smallest.

3. Final Answer:

   Perform an in-order traversal and stop when the kth element is reached. -> Option A

4. Quick Check:

   In-order traversal = sorted order [OK]

Hint: In-order traversal visits nodes in ascending order [OK]

Common Mistakes:

• Using level-order which is not sorted
• Traversing only right subtree misses smaller values
• Sorting after pre-order wastes time