Validate if Tree is BST in DSA C# - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does BST stand for and what is its key property?

BST stands for Binary Search Tree. Its key property is that for every node, all nodes in its left subtree have smaller values, and all nodes in its right subtree have larger values.

beginner

What is the main idea to check if a binary tree is a BST?

The main idea is to verify that every node's value lies within a valid range, which is updated as we traverse the tree, ensuring left children are less and right children are greater than the current node.

intermediate

Why can't we just check if the left child is less and right child is greater to validate BST?

Because this only checks immediate children. The BST property must hold for all descendants, so we must check the entire subtree values against valid ranges, not just direct children.

beginner

What is the time complexity of validating a BST using range limits?

The time complexity is O(n), where n is the number of nodes, because we visit each node once to check its value against the valid range.

beginner

Show a simple C# method signature to validate if a tree is BST.

public bool IsValidBST(TreeNode root) { /* implementation */ }

What must be true for every node in a BST?

A Left subtree nodes are smaller, right subtree nodes are larger

B Left subtree nodes are larger, right subtree nodes are smaller

C All nodes have equal values

D No children nodes

In a BST, all nodes in the left subtree must be smaller and all nodes in the right subtree must be larger than the current node.

Which approach correctly validates a BST?

A Check only immediate children values

B Count nodes in left and right subtree

C Check if tree is balanced

D Check node values against a valid range during traversal

Validating BST requires checking each node's value against a valid range that updates as we traverse the tree.

What is the time complexity of validating a BST by checking all nodes once?

A O(log n)

B O(n)

C O(1)

D O(n^2)

Each node is visited once, so the time complexity is O(n).

If a node violates the BST property, what should the validation function do?

A Return false immediately

B Return true

C Ignore and continue

D Delete the node

If any node violates the BST property, the tree is not a BST, so the function should return false immediately.

Which data structure is commonly used to represent nodes in a BST in C#?

A List

B Array

C TreeNode class with left and right children

D Dictionary

A TreeNode class with left and right child references is commonly used to represent BST nodes.

Explain how to validate if a binary tree is a BST using range limits.

Think about how the valid range changes as you move down the tree.

Describe why checking only immediate children is not enough to validate a BST.

Consider a node's left child is valid but a grandchild violates the BST property.

Practice

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

easy

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

B. Each node has exactly two children

C. All nodes have unique values but no order is required

D. The tree is always balanced

Solution

1. Step 1: Understand BST property

   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 definition

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

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Confusing BST with balanced tree
• Thinking all nodes must have two children
• Ignoring the ordering property

2. Which of the following C# method signatures is correct to check if a binary tree is a BST?

easy

A. bool IsBST(Node root, int min, int max)

B. void IsBST(Node root)

C. int IsBST(Node root)

D. bool IsBST(Node root, Node parent)

Solution

1. Step 1: Identify parameters needed for BST validation

   To validate BST, we need to check node values within a range (min, max) as we traverse.

2. Step 2: Match method signature with required parameters

   bool IsBST(Node root, int min, int max) includes root node and min/max range, suitable for recursive BST check.

3. Final Answer:

   bool IsBST(Node root, int min, int max) -> Option A

4. Quick Check:

   Method with min/max range = bool IsBST(Node root, int min, int max) [OK]

Hint: Use min and max range parameters for BST validation [OK]

Common Mistakes:

• Using void return type instead of bool
• Missing min and max parameters for range checks
• Passing parent node instead of value ranges

3. What will be the output of the following C# code snippet?

class Node {
  public int val;
  public Node left, right;
  public Node(int v) { val = v; left = right = null; }
}

bool IsBST(Node node, int min, int max) {
  if (node == null) return true;
  if (node.val <= min || node.val >= max) return false;
  return IsBST(node.left, min, node.val) && IsBST(node.right, node.val, max);
}

var root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);
root.left.left = new Node(1);
root.left.right = new Node(12);

Console.WriteLine(IsBST(root, int.MinValue, int.MaxValue));

medium

A. True

B. False

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Trace the tree structure and BST property

   The tree has root 10, left child 5, right child 15. Left child's right child is 12, which is greater than 10 but in left subtree.

2. Step 2: Check BST validation logic

   IsBST will fail because 12 is in left subtree of 10 but greater than 10, violating BST rules.

3. Final Answer:

   False -> Option B

4. Quick Check:

   Node 12 in left subtree breaks BST = False [OK]

Hint: Check if left subtree nodes are less than root [OK]

Common Mistakes:

• Assuming all nodes are valid without checking subtree ranges
• Ignoring node 12's position violating BST
• Confusing left and right subtree values

4. Identify the bug in the following C# code that checks if a binary tree is a BST:

bool IsBST(Node node) {
  if (node == null) return true;
  if (node.left != null && node.left.val > node.val) return false;
  if (node.right != null && node.right.val < node.val) return false;
  return IsBST(node.left) && IsBST(node.right);
}

medium

A. It should return int instead of bool

B. It uses incorrect comparison operators

C. It does not handle null nodes correctly

D. It does not check the entire subtree values, only immediate children

Solution

1. Step 1: Analyze the checks performed

   The code only compares node values with immediate left and right children, not entire subtrees.

2. Step 2: Understand BST validation requirement

   BST requires all nodes in left subtree < node, and all in right subtree > node, not just immediate children.

3. Final Answer:

   It does not check the entire subtree values, only immediate children -> Option D

4. Quick Check:

   Check full subtree values, not just children [OK]

Hint: Check full subtree ranges, not just direct children [OK]

Common Mistakes:

• Only comparing immediate children values
• Ignoring subtree min/max constraints
• Assuming null handling is incorrect

5. Given a binary tree, how would you modify the standard BST validation to allow duplicate values only on the right subtree?

hard

A. Change the comparison to allow node.val <= max for right subtree and node.val > min for left subtree

B. Allow duplicates only if node.val >= min for left subtree and node.val < max for right subtree

C. Allow duplicates only if node.val > min for left subtree and node.val <= max for right subtree

D. Disallow duplicates completely by strict less than and greater than comparisons

Solution

1. Step 1: Understand duplicate placement rule

   Duplicates allowed only on right subtree means left subtree nodes must be strictly less than node, right subtree nodes can be equal or greater.

2. Step 2: Adjust comparison operators accordingly

   Left subtree: node.val > min (strictly less), right subtree: node.val <= max (allow equal or greater).

3. Final Answer:

   Allow duplicates only if node.val > min for left subtree and node.val <= max for right subtree -> Option C

4. Quick Check:

   Duplicates on right subtree use <= max [OK]

Hint: Use <= for right subtree, > for left subtree to allow duplicates right only [OK]

Common Mistakes:

• Allowing duplicates on left subtree
• Using <= on left subtree incorrectly
• Disallowing duplicates completely when allowed