Check if Binary Tree is Balanced in DSA Python - Build from Scratch

Check if Binary Tree is Balanced

📖 Scenario: You are working on a program that manages a family tree. To keep the tree easy to search and update, you want to check if the tree is balanced. A balanced tree means the left and right branches of every node do not differ in height by more than one.

🎯 Goal: Build a program that creates a binary tree, sets up a helper function to check heights, then writes the logic to check if the tree is balanced.

📋 What You'll Learn

Create a binary tree using a Node class with value, left, and right attributes

Create a helper function height that returns the height of a node

Create a function is_balanced that returns True if the tree is balanced, otherwise False

Print the result of is_balanced(root) to show if the tree is balanced

💡 Why This Matters

🌍 Real World

Balanced binary trees are used in databases and file systems to keep data organized for fast searching and updating.

💼 Career

Understanding tree balance is important for software engineers working with data structures, algorithms, and performance optimization.

1

Create the Binary Tree Structure

Create a class called Node with an __init__ method that takes value, left, and right parameters. Then create a binary tree with root node value 1, left child value 2, and right child value 3. The left child should have a left child with value 4. Assign the root node to a variable called root.

DSA Python

# Define Node class and create the binary tree
# Your code here

Hint

Define the Node class first. Then create nodes inside nodes to build the tree.

2

Create a Helper Function to Find Height

Create a function called height that takes a node parameter. It should return 0 if the node is None. Otherwise, return 1 + max(height(node.left), height(node.right)).

DSA Python

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

root = Node(1, Node(2, Node(4)), Node(3))

# Define the height function below
# Your code here

Hint

Use recursion to find the height of left and right children, then add 1 for the current node.

3

Write the Function to Check if Tree is Balanced

Create a function called is_balanced that takes a node parameter. If the node is None, return True. Calculate the heights of the left and right children using the height function. If the absolute difference between these heights is more than 1, return False. Otherwise, return is_balanced(node.left) and is_balanced(node.right).

DSA Python

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

root = Node(1, Node(2, Node(4)), Node(3))

def height(node):
    if node is None:
        return 0
    return 1 + max(height(node.left), height(node.right))

# Define the is_balanced function below
# Your code here

Hint

Check the height difference at each node and recursively check left and right subtrees.

4

Print the Result of Balance Check

Write a print statement to display the result of calling is_balanced(root).

DSA Python

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

root = Node(1, Node(2, Node(4)), Node(3))

def height(node):
    if node is None:
        return 0
    return 1 + max(height(node.left), height(node.right))

def is_balanced(node):
    if node is None:
        return True
    left_height = height(node.left)
    right_height = height(node.right)
    if abs(left_height - right_height) > 1:
        return False
    return is_balanced(node.left) and is_balanced(node.right)

# Print the balance check result below
# Your code here

Hint

Use print(is_balanced(root)) to show if the tree is balanced.

Practice

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

easy

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

B. All nodes have two children

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

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 this definition exactly, while others describe different tree properties.

3. Final Answer:

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

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 of the following Python function signatures is correct to check if a binary tree is balanced?

easy

A. def is_balanced(root: List[int]) -> bool:

B. def is_balanced(root: int) -> int:

C. def is_balanced(tree: str) -> bool:

D. def is_balanced(root: Optional[TreeNode]) -> bool:

Solution

1. Step 1: Identify correct parameter type for binary tree root

   The root of a binary tree is typically a TreeNode or None, so Optional[TreeNode] is correct.

2. Step 2: Check return type for balance check

   The function should return a boolean indicating if the tree is balanced, so return type bool is correct.

3. Final Answer:

   def is_balanced(root: Optional[TreeNode]) -> bool: -> Option D

4. Quick Check:

   Function signature matches tree root and bool return [OK]

Hint: Use TreeNode or Optional[TreeNode] as input, bool as output [OK]

Common Mistakes:

• Using int or str as input instead of TreeNode
• Returning int instead of bool
• Using list instead of tree node

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

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def check_height(node):
    if not node:
        return 0
    left_height = check_height(node.left)
    right_height = check_height(node.right)
    if left_height == -1 or right_height == -1 or abs(left_height - right_height) > 1:
        return -1
    return max(left_height, right_height) + 1

def is_balanced(root):
    return check_height(root) != -1

root = TreeNode(1, TreeNode(2, TreeNode(3)), TreeNode(4))
print(is_balanced(root))

medium

A. Runtime Error

B. True

C. None

D. False

Solution

1. Step 1: Analyze tree structure

   The tree root has left child 2 with left child 3, and right child 4. Left subtree height is 2, right subtree height is 1.

2. Step 2: Check balance condition

   At root, height difference is abs(2 - 1) = 1, which is allowed. But node 2 has left child 3 and no right child, so difference is abs(1 - 0) = 1, allowed. So tree is balanced.

3. Step 3: Re-examine code logic

   Wait, the code returns -1 if difference > 1. Here difference never exceeds 1, so check_height returns height, is_balanced returns True.

4. Step 4: Re-examine the tree carefully

   Node 2 has left child 3 (height 1) and no right child (height 0), difference is 1 (allowed). Root has left subtree height 2, right subtree height 1, difference 1 (allowed). So tree is balanced.

5. Final Answer:

   True -> Option B

6. Quick Check:

   Height difference ≤ 1 means balanced = True [OK]

Hint: Check height difference at each node; -1 means unbalanced [OK]

Common Mistakes:

• Miscounting subtree heights
• Confusing -1 return as False directly
• Ignoring base case for None nodes

4. Find the bug in this code that checks if a binary tree is balanced:

def is_balanced(root):
    def height(node):
        if not node:
            return 0
        left = height(node.left)
        right = height(node.right)
        if abs(left - right) > 1:
            return False
        return max(left, right) + 1
    return height(root) != False

medium

A. Returning False instead of -1 breaks height calculation

B. Using != False instead of is not False causes error

C. Missing base case for leaf nodes

D. Not checking left and right subtree balance separately

Solution

1. Step 1: Understand return values in height function

   The height function returns integer heights, but returns False when unbalanced, mixing types.

2. Step 2: Identify impact on recursion

   Returning False breaks max(left, right) + 1 calculation because max expects integers, causing errors or wrong results.

3. Final Answer:

   Returning False instead of -1 breaks height calculation -> Option A

4. Quick Check:

   Use -1 for unbalanced, not False [OK]

Hint: Use -1 for unbalanced, not False, to keep height numeric [OK]

Common Mistakes:

• Mixing boolean and integer return types
• Incorrect final check with != False
• Not propagating unbalanced signal properly

5. Given a binary tree, which approach efficiently checks if it is balanced without recalculating heights multiple times?

hard

A. Use a bottom-up recursion returning height or -1 if unbalanced

B. Calculate height for each node separately, then check balance

C. Traverse tree twice: once for height, once for balance

D. Use breadth-first search to check balance at each level

Solution

1. Step 1: Understand naive approach drawbacks

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

2. Step 2: Identify efficient approach

   Bottom-up recursion returns height if balanced, or -1 if unbalanced, stopping early and avoiding repeated height calculations.

3. Step 3: Compare other options

   Options B and C cause repeated calculations; D (BFS) does not directly check subtree heights efficiently.

4. Final Answer:

   Use a bottom-up recursion returning height or -1 if unbalanced -> Option A

5. Quick Check:

   Bottom-up recursion = efficient balance check [OK]

Hint: Return -1 early in recursion to avoid repeated height checks [OK]

Common Mistakes:

• Using top-down approach causing repeated height calls
• Trying BFS which doesn't check subtree heights easily
• Checking balance after separate height computations