BST property and invariant in Data Structures Theory - Step-by-Step Execution

Concept Flow - BST property and invariant

Start at root node

↓

Check left subtree

↓

All left nodes < current node?

No→BST property violated

Yes↓

Check right subtree

↓

All right nodes > current node?

No→BST property violated

Yes↓

Repeat for all nodes

↓

If all checks pass

→BST property holds

The BST property means every node's left subtree has smaller values and right subtree has larger values. This must hold for every node.

Execution Sample

Data Structures Theory

Node(10)
 /      \
5       15
/ \     / \
2  7   12  20

A BST where each node's left children are smaller and right children are larger, maintaining the BST property.

Analysis Table

Step	Node Checked	Left Subtree Values	Right Subtree Values	BST Property Holds?
1	10	[5, 2, 7]	[15, 12, 20]	Yes, all left < 10 and right > 10
2	5	[2]	[7]	Yes, all left < 5 and right > 5
3	15	[12]	[20]	Yes, all left < 15 and right > 15
4	2	[]	[]	Yes, no children
5	7	[]	[]	Yes, no children
6	12	[]	[]	Yes, no children
7	20	[]	[]	Yes, no children
8	All nodes checked	-	-	BST property holds for entire tree

💡 All nodes satisfy the BST property, so the tree is a valid BST.

State Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 8
Current Node	root (10)	5	15	20	All nodes checked
Left Subtree Values	N/A	[2]	[12]	[]	N/A
Right Subtree Values	N/A	[7]	[20]	[]	N/A
BST Property Holds	Unknown	Yes	Yes	Yes	Yes

Key Insights - 3 Insights

Why do we check all nodes, not just the root?

Because the BST property must hold at every node, not just the root. The execution_table shows checks at each node (steps 1-7).

What if a left subtree node is greater than the current node?

Then the BST property is violated. The flow would go to 'BST property violated' as shown in the concept_flow diagram.

Why do leaf nodes automatically satisfy the BST property?

Leaf nodes have no children, so no left or right subtree to violate the property, as shown in steps 4-7 in the execution_table.

Visual Quiz - 3 Questions

Look at the execution_table, what is the left subtree values checked at node 5?

A [2]

B [5, 2, 7]

C []

D [7]

💡 Hint

Check row with Step 2 in execution_table under 'Left Subtree Values'

At which step does the execution_table confirm all nodes have been checked?

A Step 1

B Step 7

C Step 8

D Step 3

💡 Hint

Look for the row mentioning 'All nodes checked' in execution_table

If a node's right subtree contains a value less than the node, what happens?

A BST property holds

B BST property violated

C Check continues to next node

D Tree becomes balanced

💡 Hint

Refer to concept_flow where right subtree check fails leading to violation

Concept Snapshot

BST Property:
- For each node, all left subtree nodes < node value
- All right subtree nodes > node value
- Must hold for every node
- Ensures efficient search, insert, delete
- Violation means tree is not a BST

Full Transcript

The BST property means every node's left subtree contains only smaller values and the right subtree only larger values. This rule applies to every node in the tree. We check each node's left and right subtree values to confirm this. Leaf nodes automatically satisfy the property as they have no children. If any node violates this rule, the tree is not a BST. The execution table shows step-by-step checks for each node, confirming the property holds throughout the tree.

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