Check if Binary Tree is Balanced in DSA Rust - 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.

This helps us understand how the time grows as the tree gets bigger.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn is_balanced(root: &Option>) -> bool {
    fn height(node: &Option>) -> i32 {
        if let Some(n) = node {
            let left = height(&n.left);
            let right = height(&n.right);
            if left == -1 || right == -1 || (left - right).abs() > 1 {
                return -1;
            }
            left.max(right) + 1
        } else {
            0
        }
    }
    height(root) != -1
}
    

This code checks if a binary tree is balanced by computing heights and verifying balance at each node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls to visit each node once.
• How many times: Each node is visited once to calculate height and check balance.

How Execution Grows With Input

As the number of nodes grows, the function visits each node once.

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

Pattern observation: The work 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 visits nodes multiple times, so it must be O(n²)."

[OK] Correct: The code visits each node only once because it returns -1 early if unbalanced, avoiding repeated work.

Interview Connect

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

Self-Check

"What if the function calculated height separately for left and right subtrees without early stopping? How would the time complexity change?"

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 sorted in ascending order

Solution

1. Step 1: Understand balance condition

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

2. Step 2: Eliminate incorrect options

   All nodes having two children describes a full tree. Equal number of nodes in subtrees does not define balance. Sorted order defines a BST. None match the balance definition.

3. Final Answer:

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

4. Quick Check:

   Balance means height difference ≤ 1 [OK]

Hint: Balance means subtree heights differ by no more than 1 [OK]

Common Mistakes:

• Confusing balanced with complete or full tree
• Thinking balance means equal number of nodes
• Assuming sorted order means balanced

2. Which Rust function signature correctly represents a function that checks if a binary tree is balanced?

easy

A. fn is_balanced(root: Option) -> bool

B. fn is_balanced(root: Option<TreeNode>) -> bool

C. fn is_balanced(root: &Option) -> bool

D. fn is_balanced(root: TreeNode) -> bool

Solution

1. Step 1: Understand Rust ownership and references

   Binary trees are usually represented with Option<Box<TreeNode>>. Passing by reference (&Option<Box<TreeNode>>) avoids moving ownership.

2. Step 2: Analyze the signatures

   Passing Option<Box<TreeNode>> moves ownership. Option<TreeNode> is invalid due to infinite size. TreeNode directly ignores None roots. &Option<Box<TreeNode>> correctly borrows the tree.

3. Final Answer:

   fn is_balanced(root: &Option<Box<TreeNode>>) -> bool -> Option C

4. Quick Check:

   Use reference to Option<Box<TreeNode>> [OK]

Hint: Use &Option to avoid ownership moves [OK]

Common Mistakes:

• Passing TreeNode directly instead of boxed option
• Not using references causing ownership errors
• Using Option<TreeNode> instead of Option

3. Given the following Rust code snippet, what will be the output?

struct TreeNode {
    val: i32,
    left: Option,
    right: Option,
}

fn height(node: &Option) -> i32 {
    match node {
        None => 0,
        Some(n) => 1 + i32::max(height(&n.left), height(&n.right)),
    }
}

fn is_balanced(root: &Option) -> bool {
    match root {
        None => true,
        Some(n) => {
            let lh = height(&n.left);
            let rh = height(&n.right);
            (lh - rh).abs() <= 1 && is_balanced(&n.left) && is_balanced(&n.right)
        }
    }
}

fn main() {
    let tree = Some(Box::new(TreeNode {
        val: 1,
        left: Some(Box::new(TreeNode {
            val: 2,
            left: None,
            right: None,
        })),
        right: None,
    }));
    println!("{}", is_balanced(&tree));
}

medium

A. true

B. false

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Analyze tree structure

   The tree has root 1, left child 2, and no right child. Left subtree height = 1, right subtree height = 0.

2. Step 2: Check balance condition

   Height difference is 1 which is ≤ 1. Both left and right subtrees are leaves or None, so balanced. So overall tree is balanced.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Height difference ≤ 1 means balanced [OK]

Hint: Check subtree heights and recursive balance [OK]

Common Mistakes:

• Assuming missing right child means unbalanced
• Confusing height calculation logic
• Ignoring recursive balance checks

4. Identify the bug in this Rust function that checks if a binary tree is balanced:

fn is_balanced(root: &Option) -> bool {
    if root.is_none() {
        return true;
    }
    let node = root.as_ref().unwrap();
    let lh = height(&node.left);
    let rh = height(&node.right);
    if (lh - rh).abs() > 1 {
        return false;
    }
    is_balanced(&node.left);
    is_balanced(&node.right);
    true
}

medium

A. Using unwrap() causes panic on None

B. Incorrect height calculation function used

C. Wrong type used for root parameter

D. Missing return of recursive calls results in ignoring their results

Solution

1. Step 1: Review recursive calls

   The recursive calls to is_balanced(&node.left) and is_balanced(&node.right) are made but their results are not used or returned.

2. Step 2: Understand impact

   Ignoring recursive results means function always returns true if root is not None and height difference ≤ 1, missing unbalanced deeper nodes.

3. Final Answer:

   Missing return of recursive calls results in ignoring their results -> Option D

4. Quick Check:

   Return recursive results to propagate balance status [OK]

Hint: Always return recursive call results in boolean checks [OK]

Common Mistakes:

• Not returning recursive call results
• Assuming unwrap() always safe without check
• Confusing parameter types

5. You want to optimize the balance check to avoid repeated height calculations in Rust. Which approach is best?

hard

A. Calculate height first, then check balance separately in another traversal

B. Calculate height and balance in one recursive function returning Option<i32> height or None if unbalanced

C. Use a global variable to store heights of nodes

D. Only check balance at root node

Solution

1. Step 1: Understand repeated height calculation problem

   Calculating height separately for each node causes repeated work, increasing time complexity.

2. Step 2: Use combined recursive function

   Returning height if balanced or None if unbalanced lets us stop early and avoid repeated height calls.

3. Step 3: Analyze other options

   Separate height and balance traversals is less efficient. Global variables introduce unsafe state. Root-only check misses subtree imbalances.

4. Final Answer:

   Calculate height and balance in one recursive function returning Option<i32> height or None if unbalanced -> Option B

5. Quick Check:

   Combine height and balance check for efficiency [OK]

Hint: Return height or None to detect imbalance early [OK]

Common Mistakes:

• Doing separate height and balance traversals
• Using global variables for tree state
• Checking balance only at root