Check if Binary Tree is Balanced in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a node is null.

DSA C#

if (node == [1]) return 0;

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

Anull

B0

Cfalse

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of null for node check.

We check if the node is null to handle the base case of recursion.

2fill in blank

medium

Complete the code to get the maximum of two integers.

DSA C#

int height = Math.[1](leftHeight, rightHeight) + 1;

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

AMin

BRound

CAbs

DMax

Attempts:

3 left

💡 Hint

Common Mistakes

Using Math.Min instead of Math.Max.

We use Math.Max to get the larger height between left and right subtrees.

3fill in blank

hard

Fix the error in the condition to check balance difference.

DSA C#

if (Math.Abs(leftHeight - rightHeight) > [1]) return false;

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

A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 or 2 instead of 1.

The tree is balanced if the height difference is at most 1.

4fill in blank

hard

Fill both blanks to complete the recursive height calculation and balance check.

DSA C#

int leftHeight = [1](node.left);
int rightHeight = [2](node.right);

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

AGetHeight

BCheckBalance

CCalculateDepth

DIsBalanced

Attempts:

3 left

💡 Hint

Common Mistakes

Using balance check method instead of height method.

We recursively call GetHeight to find heights of left and right subtrees.

5fill in blank

hard

Fill all three blanks to complete the balance check method with recursive calls and return statement.

DSA C#

bool [1](TreeNode node) {
    if (node == null) return true;
    int leftHeight = GetHeight(node.left);
    int rightHeight = GetHeight(node.right);
    if (Math.Abs(leftHeight - rightHeight) > 1) return false;
    return [2](node.left) && [3](node.right);
}

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

AIsBalanced

BCheckBalance

CIsTreeBalanced

DBalancedCheck

Attempts:

3 left

💡 Hint

Common Mistakes

Using different method names for recursive calls.

The method IsBalanced recursively checks if left and right subtrees are balanced.

Practice

1. What does it mean for a binary tree to be balanced?

easy

A. The height difference between left and right subtrees of every node is at most 1

B. All nodes have two children

C. The tree has the same number of nodes on left and right

D. The tree is sorted in ascending order

Solution

1. Step 1: Understand balance condition

   A balanced binary tree means the height difference between left and right subtrees of every node is at most 1.

2. Step 2: Compare other options

   Options B, C, and D describe different properties not related to balance.

3. Final Answer:

   The height difference between left and right subtrees of every node is at most 1 -> Option A

4. Quick Check:

   Balance means height difference ≤ 1 [OK]

Hint: Balance means subtree heights differ by no more than 1 [OK]

Common Mistakes:

• Confusing balanced with complete or full tree
• Thinking balance means equal number of nodes
• Assuming sorted order means balanced

2. Which C# method signature correctly represents a function that checks if a binary tree is balanced?

easy

A. public bool IsBalanced(TreeNode root)

B. bool CheckBalance(int root)

C. public int IsBalanced(TreeNode root)

D. void IsBalanced(TreeNode root)

Solution

1. Step 1: Identify correct return type and parameters

   The function should return a boolean indicating if the tree is balanced and take the root node as input.

2. Step 2: Check method signature correctness

   public bool IsBalanced(TreeNode root) matches this: public bool IsBalanced(TreeNode root). Others have wrong return types or parameter types.

3. Final Answer:

   public bool IsBalanced(TreeNode root) -> Option A

4. Quick Check:

   Return bool, input TreeNode root [OK]

Hint: Check method returns bool and takes TreeNode root [OK]

Common Mistakes:

• Using int return type instead of bool
• Passing wrong parameter type like int
• Using void return type for a check function

3. Given the following tree structure and code, what will be the output?

class TreeNode {
  public int val;
  public TreeNode left, right;
  public TreeNode(int v) { val = v; }
}

// Tree:
//     1
//    / \
//   2   3
//  /
// 4

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

bool result = IsBalanced(root);
Console.WriteLine(result);

bool IsBalanced(TreeNode node) {
  if (node == null) return true;
  int leftHeight = Height(node.left);
  int rightHeight = Height(node.right);
  if (Math.Abs(leftHeight - rightHeight) > 1) return false;
  return IsBalanced(node.left) && IsBalanced(node.right);
}

int Height(TreeNode node) {
  if (node == null) return 0;
  return 1 + Math.Max(Height(node.left), Height(node.right));
}

medium

A. Runtime exception

B. false

C. Compilation error

D. true

Solution

1. Step 1: Calculate heights of subtrees

   Node 2 has left subtree height 1 (node 4) and right subtree height 0, difference is 1 (allowed). Node 1 has left height 2 and right height 1, difference is 1 (allowed).

2. Step 2: Check balance recursively

   Node 4 and 3 are leaves (balanced). However, node 2's left subtree is height 1 and right is 0, difference 1 (balanced). So all nodes satisfy balance condition.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Height difference ≤ 1 at all nodes = true [OK]

Hint: Check height difference at each node recursively [OK]

Common Mistakes:

• Miscounting subtree heights
• Not checking all nodes recursively
• Assuming difference > 0 means unbalanced

4. The following code is intended to check if a binary tree is balanced. What is the bug?

bool IsBalanced(TreeNode node) {
  if (node == null) return true;
  int leftHeight = Height(node.left);
  int rightHeight = Height(node.right);
  if (Math.Abs(leftHeight - rightHeight) > 1) return false;
  return IsBalanced(node.left) || IsBalanced(node.right);
}

int Height(TreeNode node) {
  if (node == null) return 0;
  return 1 + Math.Max(Height(node.left), Height(node.right));
}

medium

A. Missing null check for node

B. Using || instead of && in recursive calls

C. Height function returns wrong value

D. Incorrect base case in IsBalanced

Solution

1. Step 1: Analyze recursive condition

   The code uses || (OR) to combine left and right subtree balance checks, which returns true if either subtree is balanced.

2. Step 2: Correct logic for balance check

   Both left and right subtrees must be balanced, so && (AND) should be used instead of ||.

3. Final Answer:

   Using || instead of && in recursive calls -> Option B

4. Quick Check:

   Balance requires both subtrees balanced, use && [OK]

Hint: Use && to combine subtree balance checks [OK]

Common Mistakes:

• Using || instead of && in recursive checks
• Ignoring base case for null nodes
• Incorrect height calculation

5. You want to optimize the check for a balanced binary tree to avoid repeated height calculations. Which approach is best?

hard

A. Calculate height first, then check balance separately

B. Use BFS to check balance level by level

C. Calculate height and balance in one recursive function returning a tuple

D. Store heights in a global dictionary and update after each insertion

Solution

1. Step 1: Understand repeated height calculation problem

   Calculating height separately for each node causes repeated work, increasing time complexity.

2. Step 2: Use combined recursive approach

   Returning both height and balance status in one recursive call avoids repeated height calculations and improves efficiency.

3. Final Answer:

   Calculate height and balance in one recursive function returning a tuple -> Option C

4. Quick Check:

   Combine height and balance in one recursion for efficiency [OK]

Hint: Return height and balance together in one recursion [OK]

Common Mistakes:

• Separating height and balance checks causing repeated work
• Using BFS which doesn't directly check balance
• Using global dictionary unnecessarily complicates code