BST Inorder Successor in DSA C# - Build from Scratch

BST Inorder Successor

📖 Scenario: You are working with a Binary Search Tree (BST) that stores numbers. You want to find the next bigger number (called the inorder successor) for a given number in the tree. This is useful when you want to move step-by-step through the numbers in order.

🎯 Goal: Build a program that creates a BST, sets a target node value, finds the inorder successor of that node, and prints the successor's value or null if there is none.

📋 What You'll Learn

Create a BST with the exact nodes: 20, 9, 25, 5, 12, 11, 14

Create a variable targetValue set to 9

Implement a method InorderSuccessor that finds the inorder successor of the node with targetValue

Print the value of the inorder successor node or null if no successor exists

💡 Why This Matters

🌍 Real World

Finding the next bigger element in sorted data is useful in databases, scheduling, and navigation systems where ordered traversal is needed.

💼 Career

Understanding BSTs and inorder successor logic is important for software engineers working on search algorithms, data indexing, and system optimizations.

1

Create the BST nodes

Create a class TreeNode with an int val, and TreeNode left and right properties. Then create the BST nodes with these exact values and connections:
20 as root, 9 as left child of 20, 25 as right child of 20, 5 as left child of 9, 12 as right child of 9, 11 as left child of 12, and 14 as right child of 12.

DSA C#

// Define TreeNode class and create the BST nodes with exact connections
// Your code here

Hint

Start by defining the TreeNode class with value and left/right pointers. Then create the nodes and connect them exactly as described.

2

Set the target node value

Create an int variable called targetValue and set it to 9.

DSA C#

public class TreeNode
{
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int val=0, TreeNode left=null, TreeNode right=null)
    {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

TreeNode root = new TreeNode(20);
root.left = new TreeNode(9);
root.right = new TreeNode(25);
root.left.left = new TreeNode(5);
root.left.right = new TreeNode(12);
root.left.right.left = new TreeNode(11);
root.left.right.right = new TreeNode(14);

// Create an int variable targetValue and set it to 9
// Your code here

Hint

Just create a variable named targetValue and assign it the number 9.

3

Implement the InorderSuccessor method

Write a method InorderSuccessor that takes TreeNode root and int targetValue as parameters and returns the TreeNode that is the inorder successor of the node with targetValue. Use the BST property to find the node and its successor efficiently.

DSA C#

public class TreeNode
{
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int val=0, TreeNode left=null, TreeNode right=null)
    {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

TreeNode root = new TreeNode(20);
root.left = new TreeNode(9);
root.right = new TreeNode(25);
root.left.left = new TreeNode(5);
root.left.right = new TreeNode(12);
root.left.right.left = new TreeNode(11);
root.left.right.right = new TreeNode(14);

int targetValue = 9;

// Implement InorderSuccessor method below
// Your code here

Hint

Use a loop to find the node with targetValue. Keep track of a possible successor when moving left. If the node has a right child, the successor is the leftmost node in the right subtree.

4

Print the inorder successor value

Call the InorderSuccessor method with root and targetValue. Print the successor's val if it exists, otherwise print null.

DSA C#

public class TreeNode
{
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode(int val=0, TreeNode left=null, TreeNode right=null)
    {
        this.val = val;
        this.left = left;
        this.right = right;
    }
}

TreeNode root = new TreeNode(20);
root.left = new TreeNode(9);
root.right = new TreeNode(25);
root.left.left = new TreeNode(5);
root.left.right = new TreeNode(12);
root.left.right.left = new TreeNode(11);
root.left.right.right = new TreeNode(14);

int targetValue = 9;

TreeNode InorderSuccessor(TreeNode root, int targetValue)
{
    TreeNode successor = null;
    TreeNode current = root;
    while (current != null)
    {
        if (targetValue  current.val)
        {
            current = current.right;
        }
        else
        {
            if (current.right != null)
            {
                current = current.right;
                while (current.left != null)
                {
                    current = current.left;
                }
                successor = current;
            }
            break;
        }
    }
    return successor;
}

// Call InorderSuccessor and print the result
// Your code here

Hint

Call the InorderSuccessor method with root and targetValue. If the result is not null, print its val. Otherwise, print null.

Practice

1. What is the inorder successor of a node in a Binary Search Tree (BST)?

easy

A. The node with the smallest key greater than the given node's key

B. The node with the largest key smaller than the given node's key

C. The root node of the BST

D. The left child of the given node

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits nodes in ascending order of their keys in a BST.

2. Step 2: Define inorder successor

   The inorder successor of a node is the next node visited after it in inorder traversal, which is the smallest key greater than the node's key.

3. Final Answer:

   The node with the smallest key greater than the given node's key -> Option A

4. Quick Check:

   Inorder successor = smallest greater key [OK]

Hint: Inorder successor is next bigger node in sorted order [OK]

Common Mistakes:

• Confusing successor with predecessor
• Thinking successor is always right child
• Assuming successor is parent node

2. Which of the following C# method signatures correctly represents a function to find the inorder successor of a BST node?

easy

A. public void InorderSuccessor(TreeNode root, TreeNode p)

B. public TreeNode InorderSuccessor(TreeNode root, TreeNode p)

C. public TreeNode FindSuccessor(TreeNode p)

D. public int InorderSuccessor(int root, int p)

Solution

1. Step 1: Check parameter types for BST nodes

   The method should accept the BST root and the target node to find successor, both as TreeNode objects.

2. Step 2: Check return type

   The method returns the successor node, so return type must be TreeNode, not int or void.

3. Final Answer:

   public TreeNode InorderSuccessor(TreeNode root, TreeNode p) -> Option B

4. Quick Check:

   Correct signature uses TreeNode parameters and returns TreeNode [OK]

Hint: Successor method needs root and target node, returns TreeNode [OK]

Common Mistakes:

• Using int instead of TreeNode for parameters
• Returning void instead of TreeNode
• Missing root parameter

3. Given the BST below and the node with value 15, what is the inorder successor?

        20
       /  \
     10    30
       \     \
       15    40

TreeNode successor = InorderSuccessor(root, node15); Console.WriteLine(successor.val);

medium

A. 40

B. 30

C. 20

D. 15

Solution

1. Step 1: Locate node 15 and find its inorder successor

   Node 15 has no right child, so successor is the lowest ancestor whose left child is also ancestor of 15, which is 20.

2. Step 2: Confirm successor value

   20 is the next bigger node after 15 in inorder traversal.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Successor of 15 is 20 [OK]

Hint: If no right child, go up to first bigger ancestor [OK]

Common Mistakes:

• Choosing right child 30 incorrectly
• Choosing node itself as successor
• Choosing 40 which is too far

4. The following code attempts to find the inorder successor of a node in a BST but returns null incorrectly. What is the bug?

public TreeNode InorderSuccessor(TreeNode root, TreeNode p) {
    if (p.right != null) {
        TreeNode curr = p.right;
        while (curr.left != null) curr = curr.left;
        return curr;
    }
    TreeNode successor = null;
    while (root != null) {
        if (p.val < root.val) {
            successor = root;
            root = root.left;
        } else {
            root = root.right;
        }
    }
    return successor;
}

medium

A. The code does not handle the case when p has no right child

B. The while loop condition should be curr.right instead of curr.left

C. The comparison p.val < root.val should be p.val <= root.val

D. The code is correct and should not return null incorrectly

Solution

1. Step 1: Analyze right subtree handling

   If p has right child, code correctly finds leftmost node in right subtree as successor.

2. Step 2: Analyze ancestor traversal

   If no right child, code moves up tree to find first ancestor greater than p, correctly updating successor.

3. Step 3: Confirm no logical errors

   Comparisons and loops are correct; code returns null only if no successor exists (p is max node).

4. Final Answer:

   The code is correct and should not return null incorrectly -> Option D

5. Quick Check:

   Code handles both cases correctly [OK]

Hint: Check both right subtree and ancestor paths for successor [OK]

Common Mistakes:

• Changing left to right in while loop incorrectly
• Using <= instead of < causing wrong ancestor selection
• Assuming code misses no-right-child case

5. You have a BST and want to find the inorder successor of the node with value 50. The BST is:

          40
         /  \
       20    60
      /  \     \
    10   30    70
                 \
                 80

If the node with value 50 does not exist in the tree, what should your inorder successor function return?

hard

A. null, because the node is not in the BST

B. The smallest node greater than 50, which is 60

C. The largest node smaller than 50, which is 40

D. The root node, which is 40

Solution

1. Step 1: Confirm node presence in BST

   Node with value 50 is not present in the BST, so no direct successor can be found.

2. Step 2: Understand successor function behavior

   Successor function requires the node to find its successor; if node not found, function should return null or indicate absence.

3. Final Answer:

   null, because the node is not in the BST -> Option A

4. Quick Check:

   Node missing means no successor found [OK]

Hint: Successor only valid if node exists in BST [OK]

Common Mistakes:

• Returning closest bigger node even if node missing
• Returning root node incorrectly
• Returning predecessor instead of null