Height of Binary Tree in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the height of the binary tree.

DSA Java

public int height(Node root) {
    if (root == null) {
        return [1];
    }
    return 1 + Math.max(height(root.left), height(root.right));
}

Drag options to blanks, or click blank then click option'

A0

B1

C-1

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 1 instead of 0 for null nodes.

Returning null which causes errors.

The height of an empty tree is 0, so when the node is null, we return 0.

2fill in blank

medium

Complete the code to calculate the height by comparing left and right subtree heights.

DSA Java

return 1 + Math.[1](height(root.left), height(root.right));

Drag options to blanks, or click blank then click option'

Amax

Babs

Csum

Dmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.min which gives the smaller height.

Using sum which is incorrect for height calculation.

The height of a node is 1 plus the maximum height of its left and right subtrees.

3fill in blank

hard

Fix the error in the base case to correctly handle empty nodes.

DSA Java

if (root == [1]) {
    return 0;
}

Drag options to blanks, or click blank then click option'

Afalse

B0

Cnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or false instead of null.

Using 'null' as a string instead of the keyword.

The base case checks if the node is null, not zero or false.

4fill in blank

hard

Fill both blanks to complete the recursive height calculation.

DSA Java

int leftHeight = height(root.[1]);
int rightHeight = height(root.[2]);

Drag options to blanks, or click blank then click option'

Aleft

Bright

Cparent

Dchild

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'parent' or 'child' which are not valid node references.

Mixing up left and right.

We recursively find the height of the left and right children of the current node.

5fill in blank

hard

Fill all three blanks to complete the height method with base case and recursive step.

DSA Java

public int height(Node [1]) {
    if ([2] == null) {
        return [3];
    }
    return 1 + Math.max(height([2].left), height([2].right));
}

Drag options to blanks, or click blank then click option'

Aroot

Bnode

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently.

Returning 1 instead of 0 for null nodes.

The method takes a node parameter. The base case checks if the node is null and returns 0. Then it recursively calculates height.

Practice

1. What is the height of a binary tree with only one node (the root)?

easy

A. Undefined

B. 1

C. 2

D. 0

Solution

1. Step 1: Understand height definition

   The height of a binary tree is the number of edges on the longest path from the root to a leaf.

2. Step 2: Apply to single node tree

   With only one node (root), the longest path has length 0 edges.

3. Final Answer:

   0 -> Option D

4. Quick Check:

   Single node tree height = 0 [OK]

Hint: Height counts edges on longest root-to-leaf path [OK]

Common Mistakes:

• Thinking height is 1 for single node
• Confusing height with number of nodes
• Assuming height undefined for single node

2. Which of the following Java method signatures correctly defines a recursive method to find the height of a binary tree node?

easy

A. public int height(Node root)

B. int height(Node root) void

C. public void height(Node root) returns int

D. height(Node root) int public

Solution

1. Step 1: Check Java method syntax

   Java methods must specify access modifier, return type, method name, and parameters correctly.

2. Step 2: Validate each option

   public int height(Node root) follows correct syntax: public int height(Node root). Others have syntax errors.

3. Final Answer:

   public int height(Node root) -> Option A

4. Quick Check:

   Correct Java method signature = public int height(Node root) [OK]

Hint: Java method: access + return type + name + parameters [OK]

Common Mistakes:

• Wrong order of return type and name
• Missing return type
• Using void with return value

3. What will be the output of the following Java code snippet?

class Node {
  int data;
  Node left, right;
  Node(int d) { data = d; left = right = null; }
}

public class Test {
  public static int height(Node root) {
    if (root == null) return 0;
    int leftHeight = height(root.left);
    int rightHeight = height(root.right);
    return Math.max(leftHeight, rightHeight) + 1;
  }

  public static void main(String[] args) {
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    System.out.println(height(root));
  }
}

medium

A. 1

B. 3

C. 2

D. 4

Solution

1. Step 1: Visualize the tree structure

   The tree has root 1, left child 2, right child 3, and left child of 2 is 4. The longest path is 1->2->4.

2. Step 2: Calculate height recursively

   Height is max depth: left subtree height = 2 (nodes 2->4), right subtree height = 1 (node 3), so total height = max(2,1)+1 = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Longest path length = 3 [OK]

Hint: Height = longest root-to-leaf path node count [OK]

Common Mistakes:

• Counting edges instead of nodes
• Ignoring null child base case
• Misreading tree structure

4. Identify the error in the following Java method to find the height of a binary tree:

public int height(Node root) {
  if (root == null) {
    return -1;
  }
  int leftHeight = height(root.left);
  int rightHeight = height(root.right);
  return Math.max(leftHeight, rightHeight);
}

medium

A. The method should not use recursion.

B. The method should return the sum of left and right heights.

C. The base case should return 0, not -1.

D. The method should return the minimum of left and right heights.

Solution

1. Step 1: Check base case return value

   Height of empty tree (null) should be 0 or consistent with counting edges. Returning -1 causes incorrect height calculation.

2. Step 2: Check return statement

   Method returns max of left and right heights but misses adding 1 for current node, so height is underestimated.

3. Final Answer:

   The base case should return 0, not -1. -> Option C

4. Quick Check:

   Base case must return 0 for correct height [OK]

Hint: Base case return 0, add 1 after max height [OK]

Common Mistakes:

• Returning -1 in base case
• Not adding 1 to max height
• Returning sum instead of max

5. Given a binary tree where each node stores an integer, write a method to find the height of the tree and then print all nodes at the maximum height level. Which approach correctly combines these tasks?

hard

A. First find height recursively, then use a level-order traversal to print nodes at that height.

B. Print nodes during height calculation without storing height separately.

C. Use in-order traversal to print nodes at maximum height level directly.

D. Calculate height by counting nodes and print nodes at root only.

Solution

1. Step 1: Find height recursively

   Use a recursive method to find the height of the tree, which gives the maximum depth level.

2. Step 2: Use level-order traversal

   Perform a breadth-first traversal (level-order) and print nodes only at the level equal to the height found.

3. Final Answer:

   First find height recursively, then use a level-order traversal to print nodes at that height. -> Option A

4. Quick Check:

   Separate height calculation + level-order print = First find height recursively, then use a level-order traversal to print nodes at that height. [OK]

Hint: Find height first, then print nodes at that level [OK]

Common Mistakes:

• Trying to print during height recursion
• Using in-order traversal for level nodes
• Printing root only ignoring height