Check if Binary Tree is Balanced in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

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

beginner

Why is checking the height difference important in a balanced binary tree?

Because a large height difference means one side is much deeper, which can cause inefficient operations like search or insert.

intermediate

What is the time complexity of a naive approach that checks balance by computing height separately for each node?

The naive approach has O(n²) time complexity because height is recalculated for each node leading to repeated work.

intermediate

How can we optimize the balance check to run in O(n) time?

By computing height and balance status together in one recursive function, returning -1 if unbalanced to stop early.

beginner

In Go, what is a simple way to represent a binary tree node?

Using a struct with an integer value and pointers to left and right child nodes, e.g., type TreeNode struct { Val int; Left, Right *TreeNode }.

What is the maximum allowed height difference between left and right subtrees for a balanced binary tree node?

A 2

B 1

C 0

D 3

A balanced binary tree node must have left and right subtree heights differing by at most 1.

Which approach improves the balance check to O(n) time complexity?

A Use a recursive function that returns height and balance status together

B Use iterative BFS traversal

C Traverse only left subtree

D Calculate height separately for each node

Returning height and balance status together avoids repeated height calculations, making it O(n).

In Go, what type is used to represent child nodes in a binary tree struct?

A string

B int

C *TreeNode

D bool

Child nodes are pointers to TreeNode structs, allowing recursive tree structures.

If a node's left subtree height is 3 and right subtree height is 5, is the node balanced?

A No

B Yes

C Only if left subtree is full

D Only if right subtree is full

Height difference is 2, which is more than 1, so the node is not balanced.

What does a recursive function return to indicate an unbalanced subtree in the optimized approach?

A 0

B 1

C nil

D -1

Returning -1 signals that the subtree is unbalanced, allowing early termination.

Explain how to check if a binary tree is balanced using a single recursive function.

Think about combining height calculation and balance check in one step.

Describe the difference between the naive and optimized approaches to check if a binary tree is balanced.

Focus on how many times height is computed.

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