Check if Binary Tree is Balanced in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the TreeNode struct with integer value and pointers to left and right children.

DSA Go

type TreeNode struct {
    Val   int
    Left  *[1]
    Right *TreeNode
}

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

ANode

BTreeNode

CTree

DNodeTree

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different struct name for the child pointers.

Forgetting the pointer * before the struct type.

The left child pointer must be of type *TreeNode to correctly reference the same struct type.

2fill in blank

medium

Complete the code to calculate the height of a binary tree node recursively.

DSA Go

func height(root *TreeNode) int {
    if root == nil {
        return 0
    }
    leftHeight := height(root.Left)
    rightHeight := height(root.[1])
    if leftHeight > rightHeight {
        return leftHeight + 1
    }
    return rightHeight + 1
}

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

ARoot

Bleft

Crightt

DRight

Attempts:

3 left

💡 Hint

Common Mistakes

Using lowercase 'right' which does not match the struct field.

Typo like 'rightt' or 'left' instead of 'Right'.

The right child of the node is accessed by root.Right with capital R matching the struct field.

3fill in blank

hard

Fix the error in the code that checks if the binary tree is balanced by comparing subtree heights.

DSA Go

func isBalanced(root *TreeNode) bool {
    if root == nil {
        return true
    }
    leftHeight := height(root.Left)
    rightHeight := height(root.Right)
    if abs(leftHeight - rightHeight) > [1] {
        return false
    }
    return isBalanced(root.Left) && isBalanced(root.Right)
}

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

A1

B0

C2

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 which is too strict and rejects balanced trees.

Using negative numbers which are invalid for height difference.

A binary tree is balanced if the height difference between left and right subtrees is at most 1.

4fill in blank

hard

Fill both blanks to implement the absolute value function used in balance checking.

DSA Go

func abs(x [1]) [2] {
    if x < 0 {
        return -x
    }
    return x
}

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

Aint

Bfloat64

Cstring

Dbool

Attempts:

3 left

💡 Hint

Common Mistakes

Using float64 which is for decimal numbers.

Using string or bool which are invalid types here.

The abs function takes and returns an int because tree heights are integers.

5fill in blank

hard

Fill all three blanks to optimize balance check by returning height and balance status together.

DSA Go

func checkBalance(root *TreeNode) ([1], [2]) {
    if root == nil {
        return 0, true
    }
    leftHeight, leftBalanced := checkBalance(root.Left)
    rightHeight, rightBalanced := checkBalance(root.[3])
    balanced := leftBalanced && rightBalanced && abs(leftHeight - rightHeight) <= 1
    height := leftHeight + 1
    if rightHeight + 1 > height {
        height = rightHeight + 1
    }
    return height, balanced
}

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

Aint

Bbool

CRight

Dleft

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping return types or using wrong child field name.

Using lowercase 'right' instead of 'Right'.

The function returns height as int and balance status as bool; the right child is accessed by root.Right.

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 a complete binary tree

Solution

1. Step 1: Understand the balance condition

   A balanced binary tree means for every node, the height difference between its left and right subtree is at most 1.

2. Step 2: Compare options with definition

   The height difference between left and right subtrees of every node is at most 1 matches the definition exactly. Other options describe different tree properties.

3. Final Answer:

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

4. Quick Check:

   Balanced tree = height difference ≤ 1 [OK]

Hint: Balance means height difference ≤ 1 at every node [OK]

Common Mistakes:

• Confusing balanced with complete tree
• Thinking balanced means equal number of nodes
• Assuming all nodes must have two children

2. Which Go function signature correctly represents a function to check if a binary tree is balanced?

easy

A. func isBalanced(root *TreeNode) bool

B. func isBalanced(root TreeNode) int

C. func checkBalance(root *TreeNode) string

D. func balanced(root *TreeNode) void

Solution

1. Step 1: Identify correct parameter and return types

   The function should take a pointer to the root node and return a boolean indicating balanced or not.

2. Step 2: Match options with correct signature

   func isBalanced(root *TreeNode) bool matches: input is *TreeNode pointer, output is bool. Others have wrong return types or parameter types.

3. Final Answer:

   func isBalanced(root *TreeNode) bool -> Option A

4. Quick Check:

   Input pointer, output bool = correct signature [OK]

Hint: Check function returns bool and takes *TreeNode pointer [OK]

Common Mistakes:

• Passing TreeNode by value instead of pointer
• Returning int or string instead of bool
• Using void (no return) for a check function

3. What will the output be for the following Go code snippet that checks if a binary tree is balanced?

package main
import "fmt"

type TreeNode struct {
    Val int
    Left, Right *TreeNode
}

func height(root *TreeNode) int {
    if root == nil {
        return 0
    }
    leftHeight := height(root.Left)
    rightHeight := height(root.Right)
    if leftHeight == -1 || rightHeight == -1 || abs(leftHeight-rightHeight) > 1 {
        return -1
    }
    if leftHeight > rightHeight {
        return leftHeight + 1
    }
    return rightHeight + 1
}

func isBalanced(root *TreeNode) bool {
    return height(root) != -1
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

func main() {
    root := &TreeNode{Val: 1}
    root.Left = &TreeNode{Val: 2}
    root.Right = &TreeNode{Val: 3}
    root.Left.Left = &TreeNode{Val: 4}
    root.Left.Left.Left = &TreeNode{Val: 5}
    fmt.Println(isBalanced(root))
}

medium

A. Compilation error

B. true

C. false

D. Runtime panic

Solution

1. Step 1: Analyze tree structure and height function

   The tree is left-heavy: left subtree has height 3, right subtree height 1 at root. Difference is 2 > 1, so height returns -1.

2. Step 2: Check isBalanced return value

   isBalanced returns height(root) != -1, which is false here because height returned -1.

3. Final Answer:

   false -> Option C

4. Quick Check:

   Height difference > 1 means balanced = false [OK]

Hint: Height returns -1 if unbalanced; check != -1 for balanced [OK]

Common Mistakes:

• Assuming tree is balanced without checking height difference
• Confusing return values of height function
• Ignoring left-heavy subtree depth

4. Identify the bug in this Go function that checks if a binary tree is balanced:

func isBalanced(root *TreeNode) bool {
    if root == nil {
        return false
    }
    leftHeight := height(root.Left)
    rightHeight := height(root.Right)
    if abs(leftHeight - rightHeight) > 1 {
        return false
    }
    return isBalanced(root.Left) && isBalanced(root.Right)
}

medium

A. Uses incorrect logical operator in return

B. Returns false for empty tree instead of true

C. Missing base case for leaf nodes

D. Does not check height difference correctly

Solution

1. Step 1: Check base case for empty tree

   Function returns false if root is nil, but empty tree is balanced, so should return true.

2. Step 2: Verify other logic correctness

   Height difference check and recursive calls are correct. Only base case return is wrong.

3. Final Answer:

   Returns false for empty tree instead of true -> Option B

4. Quick Check:

   Empty tree is balanced, so base case must return true [OK]

Hint: Empty tree is balanced; base case must return true [OK]

Common Mistakes:

• Returning false for nil root
• Not checking both subtrees recursively
• Confusing logical AND with OR in recursion

5. Given a binary tree, how can you efficiently check if it is balanced without recalculating heights multiple times?

hard

A. Calculate height of left and right subtrees separately for each node

B. Check balance only at the root node

C. Use breadth-first traversal to count nodes at each level

D. Use a bottom-up recursion that returns height or -1 if unbalanced

Solution

1. Step 1: Understand naive approach inefficiency

   Calculating height separately for each node causes repeated work, leading to O(n²) time.

2. Step 2: Use bottom-up recursion for efficiency

   Return height if balanced, else -1 to stop early. This avoids repeated height calculations, achieving O(n) time.

3. Final Answer:

   Use a bottom-up recursion that returns height or -1 if unbalanced -> Option D

4. Quick Check:

   Bottom-up recursion avoids repeated height calls [OK]

Hint: Return -1 on unbalanced during height calc to stop early [OK]

Common Mistakes:

• Recalculating heights multiple times causing slow code
• Checking balance only at root misses subtree imbalance
• Using BFS which doesn't help with height difference