Check if Binary Tree is Balanced in DSA Python - Time & Space Complexity

Time Complexity: Check if Binary Tree is Balanced

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a binary tree is balanced.

Specifically, how the time grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

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

This code checks if a binary tree is balanced by measuring heights and verifying the difference is not more than one at any node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Each node is visited exactly one time.

How Execution Grows With Input

As the tree grows, the function visits every node once to check heights.

Input Size (n)	Approx. Operations
10	About 10 visits
100	About 100 visits
1000	About 1000 visits

Pattern observation: The number of operations grows directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to check balance grows linearly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "The function checks each node multiple times, so it is O(n²)."

[OK] Correct: The code uses a single pass recursion that returns height and balance status together, so each node is visited only once.

Interview Connect

Understanding this linear time check shows you can analyze recursive tree problems efficiently, a key skill in many coding challenges.

Self-Check

"What if the function separately calculated height and balance in two passes? How would the time complexity change?"

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