BST Inorder Successor in DSA Java - Build from Scratch

BST Inorder Successor

📖 Scenario: You are working with a phone contact list stored as a Binary Search Tree (BST). Each contact has a unique phone number as the key. You want to find the next contact in the sorted order after a given contact.

🎯 Goal: Build a program to find the inorder successor of a given node in a BST. The inorder successor is the node with the smallest key greater than the given node's key.

📋 What You'll Learn

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

Create a variable called target pointing to the node with key 9

Implement a method inorderSuccessor that finds the inorder successor of target

Print the key of the inorder successor node

💡 Why This Matters

🌍 Real World

Finding the next contact in a sorted phonebook or database stored as a BST helps in quick navigation and search.

💼 Career

Understanding BST traversal and successor/predecessor logic is important for software roles involving data structures, databases, and system design.

1

Create the BST nodes

Create a class Node with integer key, and left and right child nodes. Then create the BST with nodes having keys 20, 9, 25, 5, 12, 11, and 14 arranged as a BST.

DSA Java

// Define Node class and create BST nodes with keys 20, 9, 25, 5, 12, 11, 14
// Your code here

Hint

Start by defining the Node class with key, left, and right. Then create the root node with key 20 and add children to build the BST.

2

Set the target node

Create a variable called target that points to the node with key 9 in the BST.

DSA Java

class Node {
    int key;
    Node left, right;

    Node(int item) {
        key = item;
        left = right = null;
    }
}

public class Main {
    public static void main(String[] args) {
        Node root = new Node(20);
        root.left = new Node(9);
        root.right = new Node(25);
        root.left.left = new Node(5);
        root.left.right = new Node(12);
        root.left.right.left = new Node(11);
        root.left.right.right = new Node(14);

        // Create target node pointing to node with key 9
        // Your code here
    }
}

Hint

Assign target to root.left because that node has key 9.

3

Implement inorderSuccessor method

Implement a static method inorderSuccessor that takes Node root and Node target and returns the inorder successor node. Use the BST property and handle the case when target has a right subtree or not.

DSA Java

class Node {
    int key;
    Node left, right;

    Node(int item) {
        key = item;
        left = right = null;
    }
}

public class Main {
    public static Node inorderSuccessor(Node root, Node target) {
        // Implement inorderSuccessor method here
        // Your code here
    }

    public static void main(String[] args) {
        Node root = new Node(20);
        root.left = new Node(9);
        root.right = new Node(25);
        root.left.left = new Node(5);
        root.left.right = new Node(12);
        root.left.right.left = new Node(11);
        root.left.right.right = new Node(14);

        Node target = root.left; // node with key 9
    }
}

Hint

Check if target has a right child. If yes, the successor is the leftmost node in the right subtree. Otherwise, traverse from root to find the lowest ancestor whose left child is also an ancestor of target.

4

Print the inorder successor key

Call inorderSuccessor with root and target and print the key of the returned node.

DSA Java

class Node {
    int key;
    Node left, right;

    Node(int item) {
        key = item;
        left = right = null;
    }
}

public class Main {
    public static Node inorderSuccessor(Node root, Node target) {
        if (target.right != null) {
            Node curr = target.right;
            while (curr.left != null) {
                curr = curr.left;
            }
            return curr;
        }

        Node successor = null;
        Node curr = root;
        while (curr != null) {
            if (target.key  curr.key) {
                curr = curr.right;
            } else {
                break;
            }
        }
        return successor;
    }

    public static void main(String[] args) {
        Node root = new Node(20);
        root.left = new Node(9);
        root.right = new Node(25);
        root.left.left = new Node(5);
        root.left.right = new Node(12);
        root.left.right.left = new Node(11);
        root.left.right.right = new Node(14);

        Node target = root.left; // node with key 9

        // Call inorderSuccessor and print the key
        // Your code here
    }
}

Hint

Call inorderSuccessor(root, target) and print the key of the returned node. If no successor exists, print a message.

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 node greater than it.

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 Java method signatures correctly represents a function to find the inorder successor in a BST node class?

easy

A. public void inorderSuccessor(Node p)

B. public Node inorderSuccessor(Node root, Node p)

C. public int inorderSuccessor(Node root)

D. public Node inorderSuccessor(int key)

Solution

1. Step 1: Identify required inputs and outputs

   To find inorder successor, we need the BST root and the target node; output is the successor node.

2. Step 2: Match method signature

   public Node inorderSuccessor(Node root, Node p) takes root and node, returns Node, matching requirements.

3. Final Answer:

   public Node inorderSuccessor(Node root, Node p) -> Option B

4. Quick Check:

   Input root and node, output node [OK]

Hint: Method needs root and target node, returns successor node [OK]

Common Mistakes:

• Missing root parameter
• Returning int instead of Node
• Using void return type

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

       20
      /  \
    10    30
      \
      15

medium

A. 20

B. 15

C. 30

D. 10

Solution

1. Step 1: Locate node 15 and check right subtree

   Node 15 has no right child, so successor is ancestor with greater value.

2. Step 2: Find lowest ancestor greater than 15

   Ancestor 20 is the next bigger node after 15.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Next bigger after 15 is 20 [OK]

Hint: If no right child, successor is nearest bigger ancestor [OK]

Common Mistakes:

• Choosing 30 which is too big
• Choosing 10 which is smaller
• Choosing the node itself

4. Consider this buggy Java code snippet to find the inorder successor of a node in BST:

public Node inorderSuccessor(Node root, Node p) {
    if (p.right != null) {
        Node curr = p.right;
        while (curr.left != null) {
            curr = curr.left;
        }
        return curr;
    }
    Node succ = null;
    while (root != null) {
        if (p.val < root.val) {
            succ = root;
            root = root.left;
        } else {
            root = root.right;
        }
    }
    return succ;
}

What is the main bug in this code?

medium

A. It uses wrong comparison operator in while loop

B. It incorrectly updates successor when p.val >= root.val

C. It does not handle the case when p has no right child

D. It may return null even if successor exists

Solution

1. Step 1: Analyze right subtree handling

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

2. Step 2: Analyze ancestor search loop

   Loop updates successor when p.val < root.val, else moves right. This is correct logic.

3. Step 3: Identify potential null return

   If p is the largest node, no successor exists, so null return is valid. Otherwise, code works.

4. Final Answer:

   It may return null even if successor exists -> Option D

5. Quick Check:

   Null return possible if logic misses successor [OK]

Hint: Check if successor can be missed when traversing ancestors [OK]

Common Mistakes:

• Assuming null means no successor always
• Ignoring right subtree case
• Misreading comparison operators

5. You have a BST and want to find the inorder successor of node with value 25. The BST is:

        40
       /  \
     20    60
    /  \  /  \
  10  30 50  70
       \
       25

Which node is the inorder successor of 25 and why?

hard

A. 30, because it is the smallest node greater than 25 in the right subtree

B. 50, because it is the leftmost node in the right subtree of 25

C. 40, because 25 has no right child and 40 is the nearest bigger ancestor

D. 20, because it is the parent of 25

Solution

1. Step 1: Check if node 25 has right child

   Node 25 has no right child, so successor is ancestor or next bigger node.

2. Step 2: Find nearest ancestor greater than 25

   Node 40 is the nearest ancestor greater than 25, so it is the inorder successor.

3. Final Answer:

   40, because 25 has no right child and 40 is the nearest bigger ancestor -> Option C

4. Quick Check:

   Next bigger node after 25 is 40 [OK]

Hint: Successor is leftmost node in right subtree or nearest bigger ancestor [OK]

Common Mistakes:

• Choosing 30 which is smaller than 25
• Choosing 50 which is too far
• Choosing parent 20 which is smaller