Create a Binary Tree Manually in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Manual Binary Tree Creation

What is the printed output of the binary tree after manually creating nodes and linking them as shown in the code?

DSA Java

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

public class BinaryTree {
    public static void main(String[] args) {
        Node root = new Node(10);
        root.left = new Node(5);
        root.right = new Node(15);
        root.left.left = new Node(3);
        root.left.right = new Node(7);
        root.right.right = new Node(18);

        printInOrder(root);
    }

    static void printInOrder(Node node) {
        if (node == null) return;
        printInOrder(node.left);
        System.out.print(node.value + " -> ");
        printInOrder(node.right);
    }
}

A 5 -> 3 -> 7 -> 10 -> 15 -> 18 ->

B 10 -> 5 -> 3 -> 7 -> 15 -> 18 ->

C 3 -> 7 -> 5 -> 10 -> 15 -> 18 ->

D 3 -> 5 -> 7 -> 10 -> 15 -> 18 ->

💡 Hint

In-order traversal prints left subtree, then node, then right subtree.

The code performs an in-order traversal of the binary tree, which visits nodes in ascending order for a binary search tree. The nodes are printed as 3, 5, 7, 10, 15, 18 with arrows.

❓ Predict Output

intermediate

Output after Adding a New Node

After adding a new node with value 6 as the left child of node 7, what is the in-order traversal output?

DSA Java

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

public class BinaryTree {
    public static void main(String[] args) {
        Node root = new Node(10);
        root.left = new Node(5);
        root.right = new Node(15);
        root.left.left = new Node(3);
        root.left.right = new Node(7);
        root.left.right.left = new Node(6);
        root.right.right = new Node(18);

        printInOrder(root);
    }

    static void printInOrder(Node node) {
        if (node == null) return;
        printInOrder(node.left);
        System.out.print(node.value + " -> ");
        printInOrder(node.right);
    }
}

A 3 -> 5 -> 6 -> 7 -> 10 -> 15 -> 18 ->

B 3 -> 6 -> 7 -> 5 -> 10 -> 15 -> 18 ->

C 3 -> 5 -> 7 -> 6 -> 10 -> 15 -> 18 ->

D 5 -> 3 -> 6 -> 7 -> 10 -> 15 -> 18 ->

💡 Hint

In-order traversal visits left subtree, node, then right subtree recursively.

The new node 6 is the left child of 7, so it appears between 5 and 7 in in-order traversal.

🔧 Debug

advanced

Identify the Error in Binary Tree Node Linking

What error will occur when running this code that manually creates a binary tree?

DSA Java

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

public class BinaryTree {
    public static void main(String[] args) {
        Node root = new Node(10);
        root.left = new Node(5);
        root.right = new Node(15);
        root.left.left = new Node(3);
        root.left.right = new Node(7);
        root.right.left.right = new Node(18);

        printInOrder(root);
    }

    static void printInOrder(Node node) {
        if (node == null) return;
        printInOrder(node.left);
        System.out.print(node.value + " -> ");
        printInOrder(node.right);
    }
}

A No error, prints: 3 -> 5 -> 7 -> 10 -> 15 -> 18 ->

B NullPointerException at root.right.left.right = new Node(18);

C SyntaxError due to missing semicolon

D ArrayIndexOutOfBoundsException

💡 Hint

Check if all intermediate nodes are initialized before accessing their children.

root.right.left is null by default, so accessing root.right.left.right causes NullPointerException.

🧠 Conceptual

advanced

Number of Nodes in a Manually Created Binary Tree

Given the following manual binary tree creation code, how many nodes does the tree contain?

DSA Java

Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.left = new Node(6);
root.right.right = new Node(7);
root.left.left.left = new Node(8);

A 6

B 7

C 8

D 9

💡 Hint

Count each new Node() call in the code.

There are 8 Node objects created and linked, so the tree has 8 nodes.

🚀 Application

expert

Manual Binary Tree Height Calculation

What is the height of the binary tree created by the following code? Height is the number of edges on the longest path from root to a leaf.

DSA Java

Node root = new Node(10);
root.left = new Node(5);
root.right = new Node(15);
root.left.left = new Node(3);
root.left.left.left = new Node(1);
root.right.right = new Node(18);
root.right.right.right = new Node(20);

A 3

B 4

C 5

D 2

💡 Hint

Count edges from root to the deepest leaf node.

The longest path from root to a leaf has 3 edges, for example, root(10) -> 5 -> 3 -> 1 or 10 -> 15 -> 18 -> 20. Thus, the height is 3.

Practice

1. What is the correct way to create a root node of a binary tree manually in Java using a Node class with an integer value?

easy

A. Node root = new Node(); root.value = 10;

B. Node root = Node(10);

C. Node root = new Node.value(10);

D. Node root = new Node(10);

Solution

1. Step 1: Understand Node object creation

   In Java, to create an object, you use the new keyword followed by the constructor call.

2. Step 2: Apply correct syntax for constructor with value

   The constructor Node(10) creates a node with value 10. So new Node(10) is correct.

3. Final Answer:

   Node root = new Node(10); -> Option D

4. Quick Check:

   Object creation with constructor = new Node(10) [OK]

Hint: Use 'new' keyword with constructor to create objects [OK]

Common Mistakes:

• Omitting 'new' keyword
• Using incorrect constructor syntax
• Trying to assign value without constructor

2. Which of the following is the correct syntax to manually link a left child node with value 5 to a root node in Java?

easy

A. root.left = new Node(5);

B. root.leftNode = new Node(5);

C. root.left = Node(5);

D. root.left = new Node.left(5);

Solution

1. Step 1: Understand node linking in binary tree

   The left child of a node is usually accessed by left field. Assigning a new node requires new Node(value).

2. Step 2: Apply correct field and constructor syntax

   So root.left = new Node(5); correctly creates and links the left child.

3. Final Answer:

   root.left = new Node(5); -> Option A

4. Quick Check:

   Link left child with new Node = root.left = new Node(5) [OK]

Hint: Use 'root.left = new Node(value)' to link left child [OK]

Common Mistakes:

• Using wrong field name like leftNode
• Omitting 'new' keyword
• Incorrect constructor call without 'new'

3. Given the following code, what is the output of an inorder traversal (left, root, right)?

class Node {
  int value;
  Node left, right;
  Node(int v) { value = v; }
}

Node root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);

void inorder(Node node) {
  if (node == null) return;
  inorder(node.left);
  System.out.print(node.value + " ");
  inorder(node.right);
}

inorder(root);

medium

A. 4 2 5 1 3

B. 1 2 3 4 5

C. 1 3 2 4 5

D. 2 4 5 1 3

Solution

1. Step 1: Understand inorder traversal order

   Inorder visits left subtree, then root, then right subtree recursively.

2. Step 2: Trace traversal on given tree

   Left subtree of root(1) is node(2) with children 4 and 5.
   Traverse left subtree of 2: 4 (no children)
   Visit 2
   Traverse right subtree of 2: 5 (no children)
   Visit root 1
   Traverse right subtree of root: 3 (no children)
   Sequence: 4 2 5 1 3

3. Final Answer:

   4 2 5 1 3 -> Option A

4. Quick Check:

   Inorder = left, root, right = 4 2 5 1 3 [OK]

Hint: Inorder = left subtree, root, then right subtree [OK]

Common Mistakes:

• Confusing preorder or postorder with inorder
• Missing left or right subtree nodes
• Printing root before left subtree

4. Identify the error in the following code that tries to create a binary tree manually:

class Node {
  int value;
  Node left, right;
  Node(int v) { value = v; }
}

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

medium

A. Node class constructor is missing

B. Cannot assign right child after root creation

C. Missing 'new' keyword when creating left child node

D. Value type should be String, not int

Solution

1. Step 1: Check node creation syntax

   Creating a new node requires the 'new' keyword before constructor call.

2. Step 2: Identify incorrect line

   Line root.left = Node(5); misses 'new' keyword, causing syntax error.

3. Final Answer:

   Missing 'new' keyword when creating left child node -> Option C

4. Quick Check:

   Object creation needs 'new' keyword [OK]

Hint: Always use 'new' to create objects in Java [OK]

Common Mistakes:

• Omitting 'new' keyword
• Confusing constructor call with method call
• Assuming automatic object creation

5. You want to manually create a binary tree with root value 7, left child 3, right child 9, and left child's right child 5. Which code snippet correctly builds this tree?

hard

A. Node root = new Node(7);
root.left = new Node(3);
root.right = new Node(9);
root.right.left = new Node(5);

B. Node root = new Node(7);
root.left = new Node(3);
root.right = new Node(9);
root.left.right = new Node(5);

C. Node root = new Node(7);
root.left = new Node(3);
root.left.left = new Node(9);
root.left.right = new Node(5);

D. Node root = new Node(7);
root.right = new Node(3);
root.left = new Node(9);
root.left.right = new Node(5);

Solution

1. Step 1: Create root and direct children

   Root is 7, left child 3, right child 9, so assign accordingly.

2. Step 2: Add left child's right child

   Left child's right child is 5, so root.left.right = new Node(5); is correct.

3. Final Answer:

   Node root = new Node(7); root.left = new Node(3); root.right = new Node(9); root.left.right = new Node(5); -> Option B

4. Quick Check:

   Assign children correctly by following tree structure [OK]

Hint: Assign children stepwise matching tree structure [OK]

Common Mistakes:

• Swapping left and right children
• Assigning child to wrong parent node
• Missing intermediate child nodes