BST Inorder Successor in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Find the inorder successor of a node in BST

What is the output of the following code that finds the inorder successor of node 15 in the BST?

DSA Java

class Node {
    int val;
    Node left, right;
    Node(int val) { this.val = val; }
}

public class BSTSuccessor {
    public static Node inorderSuccessor(Node root, Node p) {
        Node successor = null;
        while (root != null) {
            if (p.val < root.val) {
                successor = root;
                root = root.left;
            } else {
                root = root.right;
            }
        }
        return successor;
    }

    public static void main(String[] args) {
        Node root = new Node(20);
        root.left = new Node(10);
        root.right = new Node(30);
        root.left.left = new Node(5);
        root.left.right = new Node(15);
        root.right.right = new Node(40);

        Node p = root.left.right; // Node with value 15
        Node succ = inorderSuccessor(root, p);
        System.out.println(succ != null ? succ.val : "null");
    }
}

A 30

B null

C 15

D 20

💡 Hint

Remember inorder successor is the next bigger node in BST.

The inorder successor of 15 is 20 because 20 is the smallest node greater than 15 in the BST.

❓ Predict Output

intermediate

Inorder successor when node has right subtree

What will be printed by this code when finding the inorder successor of node 10?

DSA Java

class Node {
    int val;
    Node left, right;
    Node(int val) { this.val = val; }
}

public class BSTSuccessor {
    public static Node leftMost(Node node) {
        while (node.left != null) {
            node = node.left;
        }
        return node;
    }

    public static Node inorderSuccessor(Node root, Node p) {
        if (p.right != null) {
            return leftMost(p.right);
        }
        Node successor = null;
        while (root != null) {
            if (p.val < root.val) {
                successor = root;
                root = root.left;
            } else if (p.val > root.val) {
                root = root.right;
            } else {
                break;
            }
        }
        return successor;
    }

    public static void main(String[] args) {
        Node root = new Node(20);
        root.left = new Node(10);
        root.right = new Node(30);
        root.left.left = new Node(5);
        root.left.right = new Node(15);
        root.right.right = new Node(40);

        Node p = root.left; // Node with value 10
        Node succ = inorderSuccessor(root, p);
        System.out.println(succ != null ? succ.val : "null");
    }
}

A 15

B 20

C null

D 30

💡 Hint

If node has right child, successor is leftmost node in right subtree.

Node 10 has right child 15, so the inorder successor is the leftmost node in 10's right subtree, which is 15.

🧠 Conceptual

advanced

Understanding inorder successor when no right subtree

In a BST, if a node has no right subtree, how is its inorder successor determined?

A It is the parent node

B It is the lowest ancestor whose left child is also ancestor of the node

C It is the rightmost node in the left subtree

D It is always null

💡 Hint

Think about climbing up the tree to find a node bigger than current.

If no right subtree, inorder successor is the lowest ancestor for which the node lies in its left subtree.

🔧 Debug

advanced

Identify the error in inorder successor code

What error will this code produce when trying to find the inorder successor of node 25?

DSA Java

class Node {
    int val;
    Node left, right;
    Node(int val) { this.val = val; }
}

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

    public static void main(String[] args) {
        Node root = new Node(20);
        root.left = new Node(10);
        root.right = new Node(30);
        root.right.left = new Node(25);
        root.right.right = new Node(40);

        Node p = root.right.left; // Node with value 25
        Node succ = inorderSuccessor(root, p);
        System.out.println(succ != null ? succ.val : "null");
    }
}

A Prints 30

B Prints 25

C Prints null

D Runtime error NullPointerException

💡 Hint

Check the condition p.val <= root.val in the while loop.

The code prints 25, which is incorrect. The correct successor is 30, but the <= condition causes successor to be set to p itself when root reaches p.

🚀 Application

expert

Count nodes with no inorder successor in BST

Given a BST, how many nodes have no inorder successor? Consider the BST below: 20 / \ 10 30 / \ \ 5 15 40 Select the correct count.

A 3

B 2

C 1

D 4

💡 Hint

Only the maximum node in the BST has no inorder successor.

Only node 40 has no inorder successor because it is the largest node (rightmost in inorder traversal).

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