BST Inorder Predecessor in DSA Java - Build from Scratch

BST Inorder Predecessor

📖 Scenario: You are working with a Binary Search Tree (BST) that stores numbers. You want to find the inorder predecessor of a given node. The inorder predecessor is the node with the next smaller value in the BST.Think of the BST like a family tree sorted by age, and you want to find the person who is just older than a given person.

🎯 Goal: Build a Java program that creates a BST, sets a target node value, finds the inorder predecessor of that node, and prints the predecessor's value.

📋 What You'll Learn

Create a BST with exact nodes: 20, 10, 30, 5, 15, 25, 35

Set an integer variable target to 15

Implement a method inorderPredecessor to find the inorder predecessor of target

Print the value of the inorder predecessor or "No predecessor" if none exists

💡 Why This Matters

🌍 Real World

Finding the inorder predecessor is useful in database indexing and navigation systems where you need to find the previous item in sorted order.

💼 Career

Understanding BST operations like inorder predecessor is important for software engineers working with search algorithms, data retrieval, and optimization.

1

Create the BST nodes and insert them

Create a class Node with integer data, and Node references left and right. Then create a class BST with a Node root. Insert the nodes with values 20, 10, 30, 5, 15, 25, and 35 into the BST using the insert method.

DSA Java

// Define Node class
// Define BST class with insert method
// Insert nodes 20, 10, 30, 5, 15, 25, 35

Hint

Start by defining the Node class with data, left, and right. Then create the BST class with an insert method to add nodes in the correct position.

2

Set the target node value

Inside the main method, create an integer variable called target and set it to 15.

DSA Java

class Node {
    int data;
    Node left, right;

    Node(int data) {
        this.data = data;
        left = right = null;
    }
}

class BST {
    Node root;

    void insert(int data) {
        root = insertRec(root, data);
    }

    Node insertRec(Node root, int data) {
        if (root == null) {
            root = new Node(data);
            return root;
        }
        if (data  root.data) {
            root.right = insertRec(root.right, data);
        }
        return root;
    }

    public static void main(String[] args) {
        BST tree = new BST();
        tree.insert(20);
        tree.insert(10);
        tree.insert(30);
        tree.insert(5);
        tree.insert(15);
        tree.insert(25);
        tree.insert(35);
        // Create an integer variable target and set it to 15
        // Your code here
    }
}

Hint

Inside the main method, write int target = 15; to set the target value.

3

Implement the inorder predecessor method

Add a method Node inorderPredecessor(Node root, int target) in the BST class. This method should find and return the inorder predecessor node of the node with value target. Use the BST properties to find the predecessor efficiently.

DSA Java

class Node {
    int data;
    Node left, right;

    Node(int data) {
        this.data = data;
        left = right = null;
    }
}

class BST {
    Node root;

    void insert(int data) {
        root = insertRec(root, data);
    }

    Node insertRec(Node root, int data) {
        if (root == null) {
            root = new Node(data);
            return root;
        }
        if (data  root.data) {
            root.right = insertRec(root.right, data);
        }
        return root;
    }

    // Add inorderPredecessor method here
    // Your code here

    public static void main(String[] args) {
        BST tree = new BST();
        tree.insert(20);
        tree.insert(10);
        tree.insert(30);
        tree.insert(5);
        tree.insert(15);
        tree.insert(25);
        tree.insert(35);
        int target = 15;
    }
}

Hint

Use a loop starting from root. Move left if target is less or equal to current node's data. Otherwise, update predecessor to current and move right. Return predecessor at the end.

4

Print the inorder predecessor value

In the main method, call inorderPredecessor with tree.root and target. Store the result in a Node variable called pred. Then print pred.data if pred is not null, otherwise print "No predecessor".

DSA Java

class Node {
    int data;
    Node left, right;

    Node(int data) {
        this.data = data;
        left = right = null;
    }
}

class BST {
    Node root;

    void insert(int data) {
        root = insertRec(root, data);
    }

    Node insertRec(Node root, int data) {
        if (root == null) {
            root = new Node(data);
            return root;
        }
        if (data  root.data) {
            root.right = insertRec(root.right, data);
        }
        return root;
    }

    Node inorderPredecessor(Node root, int target) {
        Node predecessor = null;
        Node current = root;
        while (current != null) {
            if (target <= current.data) {
                current = current.left;
            } else {
                predecessor = current;
                current = current.right;
            }
        }
        return predecessor;
    }

    public static void main(String[] args) {
        BST tree = new BST();
        tree.insert(20);
        tree.insert(10);
        tree.insert(30);
        tree.insert(5);
        tree.insert(15);
        tree.insert(25);
        tree.insert(35);
        int target = 15;
        // Call inorderPredecessor and print the result
        // Your code here
    }
}

Hint

Call inorderPredecessor with tree.root and target. Store the result in pred. If pred is not null, print pred.data. Otherwise, print "No predecessor".

Practice

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

easy

A. The node with the largest value smaller than the given node

B. The node with the smallest value larger than the given node

C. The left child of the given node

D. The right child of the given node

Solution

1. Step 1: Understand inorder traversal order

   Inorder traversal visits nodes in ascending order for BST, so predecessor is the previous node in this order.

2. Step 2: Define inorder predecessor

   The inorder predecessor is the node with the largest value smaller than the current node's value.

3. Final Answer:

   The node with the largest value smaller than the given node -> Option A

4. Quick Check:

   Inorder predecessor = largest smaller value [OK]

Hint: Predecessor is just before current node in sorted order [OK]

Common Mistakes:

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

2. Which of the following Java method signatures correctly represents a function to find the inorder predecessor in a BST node?

easy

A. public Node findInorderSuccessor(Node root, int key)

B. public int findInorderPredecessor(int[] arr, int key)

C. public Node findInorderPredecessor(Node root, int key)

D. public void findInorderPredecessor(Node root)

Solution

1. Step 1: Identify correct parameters for BST predecessor method

   We need the BST root and the key node value to find predecessor, so method should accept Node root and int key.

2. Step 2: Check method return type and name

   Return type should be Node (the predecessor node), and method name should indicate predecessor, so 'findInorderPredecessor(Node root, int key)' fits.

3. Final Answer:

   public Node findInorderPredecessor(Node root, int key) -> Option C

4. Quick Check:

   Correct signature includes root, key, returns Node [OK]

Hint: Method needs root and key, returns Node predecessor [OK]

Common Mistakes:

• Using array instead of BST Node
• Confusing predecessor with successor method name
• Wrong return type like void or int

3. Given the BST below and the code snippet, what will be the output?

    20
   /  \
 10   30
   
Node predecessor = findInorderPredecessor(root, 20);
System.out.println(predecessor != null ? predecessor.data : "null");

Assuming findInorderPredecessor is correctly implemented.

medium

A. null

B. 10

C. 30

D. 20

Solution

1. Step 1: Identify inorder predecessor of node 20 in BST

   Inorder predecessor is the largest node smaller than 20. Left subtree of 20 has node 10, which is smaller and largest in left subtree.

2. Step 2: Confirm output of print statement

   Predecessor node data is 10, so printed output is "10".

3. Final Answer:

   10 -> Option B

4. Quick Check:

   Largest smaller node than 20 is 10 [OK]

Hint: Predecessor is max in left subtree if left exists [OK]

Common Mistakes:

• Printing null when left subtree exists
• Confusing successor with predecessor
• Printing the node itself instead of predecessor

4. Consider this buggy Java method to find the inorder predecessor in a BST:

public Node findInorderPredecessor(Node root, int key) {
    Node predecessor = null;
    while (root != null) {
        if (key < root.data) {
            root = root.left;
        } else {
            predecessor = root;
            root = root.right;
        }
    }
    return predecessor;
}

What is the main issue with this code?

medium

A. It does not handle the case when the node has a left subtree

B. It incorrectly updates predecessor when key equals root.data

C. It causes infinite loop for some inputs

D. It returns the successor instead of predecessor

Solution

1. Step 1: Analyze traversal logic for predecessor

   The code moves left if key < root.data, else moves right and updates predecessor. But it never checks left subtree of the node itself.

2. Step 2: Identify missing left subtree handling

   If the node has a left subtree, the predecessor is the maximum node in that left subtree, which this code does not check.

3. Final Answer:

   It does not handle the case when the node has a left subtree -> Option A

4. Quick Check:

   Missing left subtree max check causes wrong predecessor [OK]

Hint: Check left subtree max for predecessor if it exists [OK]

Common Mistakes:

• Ignoring left subtree max node
• Confusing predecessor with successor logic
• Assuming predecessor is always ancestor

5. You have a BST and want to find the inorder predecessor of a node with value 15. The node has no left child. Which approach correctly finds the predecessor?

hard

A. Find the maximum node in the left subtree of 15

B. Return the right child of node 15

C. Return null because no left child exists

D. Traverse from root, keep track of last node with value less than 15

Solution

1. Step 1: Understand predecessor when no left child

   If node has no left child, predecessor is one of its ancestors with value less than node's value.

2. Step 2: Use traversal from root to track predecessor

   Traverse from root to node 15, keep updating predecessor to last node with value less than 15.

3. Final Answer:

   Traverse from root, keep track of last node with value less than 15 -> Option D

4. Quick Check:

   No left child means ancestor with smaller value is predecessor [OK]

Hint: Track last smaller ancestor when no left child [OK]

Common Mistakes:

• Returning null immediately if no left child
• Trying to find max in non-existent left subtree
• Returning right child mistakenly