Check if Binary Tree is Balanced in DSA Go - Execution Trace

Concept Flow - Check if Binary Tree is Balanced

Start at root node

↓

Check left subtree height

↓

Check right subtree height

↓

Compare heights difference <= 1?

Yes No↓

Check left balanced?

↓

Check right balanced?

↓

Return true if both balanced

We check the height of left and right subtrees at each node and ensure their difference is at most 1, recursively verifying balance.

Execution Sample

DSA Go

func isBalanced(root *Node) bool {
  _, balanced := checkHeight(root)
  return balanced
}

func checkHeight(node *Node) (int, bool) {
  if node == nil { return 0, true }
  leftHeight, leftBalanced := checkHeight(node.Left)
  rightHeight, rightBalanced := checkHeight(node.Right)
  if !leftBalanced || !rightBalanced || leftHeight - rightHeight > 1 || rightHeight - leftHeight > 1 {
    return 0, false
  }
  h := leftHeight
  if rightHeight > h {
    h = rightHeight
  }
  return h + 1, true
}

This code recursively checks subtree heights and balance, returning false if any subtree is unbalanced.

Execution Table

Step	Operation	Node Visited	Left Height	Right Height	Balanced?	Visual State
1	Visit root	1	?	?	?	1 / \ 2 3
2	Visit left child	2	?	?	?	1 / \ 2 3 / 4
3	Visit left-left child	4	0	0	true	1 / \ 2 3 / 4
4	Return from node 4	4	0	0	true	1 / \ 2 3 / 4
5	Return from node 2	2	1	0	true	1 / \ 2 3 / 4
6	Visit right child	3	?	?	?	1 / \ 2 3 / 4
7	Visit right-right child	5	0	0	true	1 / \ 2 3 / 4 \ 5
8	Return from node 5	5	0	0	true	1 / \ 2 3 / 4 \ 5
9	Return from node 3	3	0	1	true	1 / \ 2 3 / 4 \ 5
10	Return from root	1	2	2	true	1 / \ 2 3 / 4 \ 5
11	Check balance condition	1	2	2	true	Balanced tree confirmed
12	End	-	-	-	-	Traversal complete, tree is balanced

💡 Traversal ends when all nodes are visited and balance conditions hold at every node.

Variable Tracker

Variable	Start	After Step 3	After Step 5	After Step 8	After Step 10	Final
node	root (1)	4	2	5	1	nil
leftHeight	?	0	1	0	2	2
rightHeight	?	?	0	1	2	2
balanced	?	true	true	true	true	true

Key Moments - 3 Insights

Why do we return height 0 and balanced true for a nil node?

Because a nil node means no subtree, so height is 0 and it's balanced by definition (see Step 3 and Step 8).

Why do we check balance after getting left and right heights?

Because balance depends on height difference and subtree balance, as shown in Steps 9 and 10 where we compare heights and balanced flags.

What happens if the height difference is more than 1?

The function returns false immediately (not shown in this balanced example), stopping further checks and marking the tree unbalanced.

Visual Quiz - 3 Questions

Look at the execution table, what is the leftHeight at Step 10?

A 1

B 0

C 2

D 3

💡 Hint

Check the 'Left Height' column at Step 10 in the execution_table.

At which step does the function confirm the tree is balanced?

A Step 11

B Step 9

C Step 5

D Step 12

💡 Hint

Look for the step where 'Balanced tree confirmed' appears in the Visual State column.

If node 3 had a right child making rightHeight 4, what would happen at Step 10?

A leftHeight would increase

B balanced would be false

C balanced would be true

D function would continue without change

💡 Hint

Recall the balance condition checks if height difference > 1, see Step 10 logic.

Concept Snapshot

Check if Binary Tree is Balanced:
- Recursively get left and right subtree heights
- If difference > 1, tree is unbalanced
- Return height and balanced status up the call stack
- Base case: nil node height=0, balanced=true
- Final result: true if all nodes balanced

Full Transcript

To check if a binary tree is balanced, we start at the root and recursively check the height of left and right subtrees. For each node, we calculate the height of its left and right children. If the difference between these heights is more than 1, the tree is not balanced. We also check if the left and right subtrees themselves are balanced. If all nodes satisfy the balance condition, the tree is balanced. The base case is when a node is nil, which means height 0 and balanced true. This process continues until all nodes are checked, returning the final balanced status.

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