BST property and invariant in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the sentence to describe the BST property: In a Binary Search Tree, the left subtree of a node contains only nodes with keys {{BLANK_1}} the node's key.

Data Structures Theory

In a Binary Search Tree, the left subtree of a node contains only nodes with keys [1] the node's key.

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

Agreater than

Bunrelated to

Cequal to

Dless than

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing left subtree keys as greater than the node's key.

Thinking keys can be equal in the left subtree.

The BST property states that all keys in the left subtree are less than the node's key.

2fill in blank

medium

Complete the sentence to describe the BST property: The right subtree of a node contains only nodes with keys {{BLANK_1}} the node's key.

Data Structures Theory

The right subtree of a node contains only nodes with keys [1] the node's key.

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

Agreater than

Bunrelated to

Cequal to

Dless than

Attempts:

3 left

💡 Hint

Common Mistakes

Assuming right subtree keys are less than the node's key.

Allowing equal keys in the right subtree.

According to the BST property, all keys in the right subtree are greater than the node's key.

3fill in blank

hard

Fix the error in this BST invariant statement: "For every node, all keys in the left subtree are {{BLANK_1}} or equal to the node's key."

Data Structures Theory

For every node, all keys in the left subtree are [1] or equal to the node's key.

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

Agreater than

Bless than

Cgreater than or equal to

Dless than or equal to

Attempts:

3 left

💡 Hint

Common Mistakes

Allowing equal keys in the left subtree.

Confusing less than with greater than.

The BST property requires keys in the left subtree to be strictly less than the node's key, not equal or greater.

4fill in blank

hard

Fill both blanks to complete the BST invariant: "For every node, all keys in the left subtree are {{BLANK_1}} the node's key, and all keys in the right subtree are {{BLANK_2}} the node's key."

Data Structures Theory

For every node, all keys in the left subtree are [1] the node's key, and all keys in the right subtree are [2] the node's key.

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

Aless than

Bequal to

Cgreater than

Dunrelated to

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up left and right subtree key comparisons.

Allowing equal keys on either side.

The BST property requires left subtree keys to be less than and right subtree keys to be greater than the node's key.

5fill in blank

hard

Fill all three blanks to complete the BST property check code snippet: "if node is not None: if node.left is not None and node.left.key {{BLANK_1}} node.key: raise ValueError('BST property violated'); if node.right is not None and node.right.key {{BLANK_2}} node.key: raise ValueError('BST property violated'); check(node.left); check(node.right)"

Data Structures Theory

if node is not None:
    if node.left is not None and node.left.key [1] node.key:
        raise ValueError('BST property violated')
    if node.right is not None and node.right.key [2] node.key:
        raise ValueError('BST property violated')
    check(node.left)
    check(node.right)

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

A>=

B<=

C<

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators that do not detect violations.

Confusing which side uses which comparison.

The left child's key must be less than the node's key (violation if left.key >= node.key). The right child's key must be greater than the node's key (violation if right.key <= node.key). The code raises an error if these conditions are violated.

Practice

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

easy

A. All nodes have exactly two children.

B. Nodes are arranged in a circular linked list.

C. Left child nodes are smaller, right child nodes are larger than the parent node.

D. Each node stores a unique key with no order.

Solution

1. Step 1: Understand BST node arrangement

   A BST arranges nodes so that left children have smaller values and right children have larger values than their parent node.

2. Step 2: Compare options with BST definition

   Left child nodes are smaller, right child nodes are larger than the parent node. correctly states this property. Other options describe different or incorrect structures.

3. Final Answer:

   Left child nodes are smaller, right child nodes are larger than the parent node. -> Option C

4. Quick Check:

   BST property = left < parent < right [OK]

Hint: Remember BST means left smaller, right larger [OK]

Common Mistakes:

• Thinking all nodes must have two children
• Confusing BST with other tree types
• Ignoring the order property in BST

2. Which of the following is the correct way to check if a node's left child maintains the BST property?

easy

A. left_child.value > node.value

B. left_child.value < node.value

C. left_child.value == node.value

D. left_child.value >= node.value

Solution

1. Step 1: Recall BST left child rule

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

2. Step 2: Match this rule to options

   left_child.value < node.value correctly states left_child.value < node.value. Other options violate this rule.

3. Final Answer:

   left_child.value < node.value -> Option B

4. Quick Check:

   Left child < parent [OK]

Hint: Left child must be smaller than parent [OK]

Common Mistakes:

• Using greater than or equal instead of less than
• Allowing equal values on left child
• Confusing left and right child rules

3. Given the BST below, what is the correct in-order traversal output?

      10
     /  \
    5    15
   / \     \
  3   7     20

medium

A. [3, 5, 7, 10, 15, 20]

B. [10, 5, 3, 7, 15, 20]

C. [20, 15, 10, 7, 5, 3]

D. [5, 3, 7, 10, 15, 20]

Solution

1. Step 1: Understand in-order traversal

   In-order traversal visits left subtree, then node, then right subtree, producing sorted order in BST.

2. Step 2: Traverse the tree in-order

   Left subtree of 10: nodes 3, 5, 7 in order; then 10; then right subtree 15, 20.

3. Final Answer:

   [3, 5, 7, 10, 15, 20] -> Option A

4. Quick Check:

   In-order traversal = sorted nodes [OK]

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

Common Mistakes:

• Confusing pre-order or post-order with in-order
• Listing nodes in insertion order
• Reversing the traversal order

4. Consider the following BST insertion code snippet. What is the error?

def insert(node, value):
    if node is null:
        return Node(value)
    if value < node.value:
        node.left = insert(node.left, value)
    else:
        node.right = insert(node.right, value)
    return node

medium

A. The code does not handle duplicate values correctly.

B. The base case for node is incorrect.

C. The recursive calls do not update the tree.

D. The function does not return the updated node.

Solution

1. Step 1: Analyze duplicate handling in insertion

   The code inserts duplicates into the right subtree without restriction, which may violate BST uniqueness if duplicates are not allowed.

2. Step 2: Check other parts of the code

   Base case and recursive updates are correct; function returns updated node properly.

3. Final Answer:

   The code does not handle duplicate values correctly. -> Option A

4. Quick Check:

   Duplicates need special handling in BST insert [OK]

Hint: Check how duplicates are treated in insertion [OK]

Common Mistakes:

• Assuming duplicates are automatically handled
• Ignoring return statements in recursion
• Confusing base case with recursive case

5. You want to verify if a binary tree is a valid BST. Which approach correctly checks the BST property for all nodes?

hard

A. Check if the tree has unique values only.

B. Check if the tree is balanced and has no cycles.

C. Check if each node's left child is smaller and right child is larger, recursively.

D. Check if all left descendants are smaller and all right descendants are larger than the node, using min and max limits.

Solution

1. Step 1: Understand BST validation requirements

   Each node must satisfy that all nodes in its left subtree are smaller and all in right subtree are larger, not just immediate children.

2. Step 2: Evaluate approaches

   Check if all left descendants are smaller and all right descendants are larger than the node, using min and max limits. uses min and max limits to ensure all descendants satisfy BST property, which is correct. Check if each node's left child is smaller and right child is larger, recursively. only checks immediate children, which is insufficient.

3. Final Answer:

   Check if all left descendants are smaller and all right descendants are larger than the node, using min and max limits. -> Option D

4. Quick Check:

   BST validation requires range checks, not just immediate children [OK]

Hint: Use min/max limits to validate BST recursively [OK]

Common Mistakes:

• Checking only immediate children values
• Confusing BST property with tree balance
• Assuming unique values guarantee BST