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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the height of the binary tree node.

DSA Python

def height(node):
    if node is None:
        return 0
    left_height = height(node.left)
    right_height = height(node.right)
    return 1 + max(left_height, [1])

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

Anode.value

Bleft_height

C0

Dright_height

Attempts:

3 left

💡 Hint

Common Mistakes

Returning min instead of max.

Using node.value instead of child's height.

The height of a node is 1 plus the maximum height of its left and right children. So we use max(left_height, right_height).

2fill in blank

medium

Complete the code to check if the current node is balanced by comparing heights of left and right subtrees.

DSA Python

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)

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

A2

B3

C1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 instead of 1, which is too strict.

Using 2 or more, which allows unbalanced trees.

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

3fill in blank

hard

Fix the error in the code to correctly check if the tree is balanced by returning a boolean.

DSA Python

def check_balanced(node):
    if node is None:
        return True
    left = check_balanced(node.left)
    right = check_balanced(node.right)
    if left is False or right is False:
        return False
    if abs(height(node.left) - height(node.right)) > [1]:
        return False
    return True

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

A1

B0

C2

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 which is too strict.

Returning None instead of False.

The height difference must not exceed 1 for the tree to be balanced.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each node's value to its height if the height is greater than 1.

DSA Python

def node_heights_dict(nodes):
    return {node.value: [1] for node in nodes if [2]

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

Aheight(node)

Bheight(node) > 1

Cheight(node) >= 1

Dnode.value > 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.value as value instead of height.

Filtering by node.value instead of height.

We map node.value to height(node) only if height(node) > 1.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase node values to their heights if height is less than or equal to 2.

DSA Python

def filtered_node_heights(nodes):
    return [1]: [2] for node in nodes if [3]

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

Anode.value.upper()

Bheight(node)

Cheight(node) <= 2

Dnode.value

Attempts:

3 left

💡 Hint

Common Mistakes

Using node.value instead of uppercase.

Filtering by node.value instead of height.

Keys are uppercase node values, values are heights, filtered by height <= 2.

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