Count Total Nodes in Binary Tree in DSA Typescript - Time & Space Complexity

Time Complexity: Count Total Nodes in Binary Tree

O(n)

Understanding Time Complexity

We want to understand how long it takes to count all nodes in a binary tree.

How does the time grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function countNodes(root: TreeNode | null): number {
  if (root === null) return 0;
  return 1 + countNodes(root.left) + countNodes(root.right);
}
    

This code counts all nodes by visiting each node once in the binary tree.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive calls visiting each node once.
• How many times: Once per node in the tree.

How Execution Grows With Input

Each node is visited once, so the work grows directly with the number of nodes.

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

Pattern observation: The time grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to count nodes grows directly with the number of nodes in the tree.

Common Mistake

[X] Wrong: "Counting nodes is faster because we only check the root or just one side."

[OK] Correct: Every node must be visited to count it, so skipping parts misses nodes and gives wrong results.

Interview Connect

Counting nodes is a basic skill that shows you understand tree traversal and recursion, which are common in many problems.

Self-Check

"What if we used an iterative approach with a stack instead of recursion? How would the time complexity change?"

Practice

1. What does the function to count total nodes in a binary tree return when the tree is empty?

easy

A. -1

B. 1

C. 0

D. null

Solution

1. Step 1: Understand the base case for an empty tree

   When the binary tree is empty, there are no nodes to count, so the count should be zero.

2. Step 2: Confirm the function's return value for empty input

   The function should return 0 to indicate no nodes exist in the tree.

3. Final Answer:

   0 -> Option C

4. Quick Check:

   Empty tree node count = 0 [OK]

Hint: Empty tree always has zero nodes [OK]

Common Mistakes:

• Returning 1 for empty tree
• Returning null instead of 0
• Confusing empty tree with leaf node

2. Which of the following TypeScript function signatures correctly defines a function to count total nodes in a binary tree node structure?

easy

A. function countNodes(root: number): number

B. function countNodes(root: any): string

C. function countNodes(root: TreeNode): void

D. function countNodes(root: TreeNode | null): number

Solution

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

   The root should be a TreeNode or null if the tree is empty.

2. Step 2: Confirm return type for count function

   The function returns a number representing total nodes, so return type must be number.

3. Final Answer:

   function countNodes(root: TreeNode | null): number -> Option D

4. Quick Check:

   Correct parameter and return types = function countNodes(root: TreeNode | null): number [OK]

Hint: Function must accept node or null and return number [OK]

Common Mistakes:

• Using wrong parameter type
• Returning void instead of number
• Returning string instead of number

3. Given the following binary tree code, what is the output of console.log(countNodes(root));?

class TreeNode {
  val: number;
  left: TreeNode | null;
  right: TreeNode | null;
  constructor(val: number) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

function countNodes(root: TreeNode | null): number {
  if (!root) return 0;
  return 1 + countNodes(root.left) + countNodes(root.right);
}

const root = new TreeNode(1);
root.left = new TreeNode(2);
root.right = new TreeNode(3);
root.left.left = new TreeNode(4);
root.left.right = new TreeNode(5);

medium

A. 5

B. 6

C. 4

D. 3

Solution

1. Step 1: Count nodes in the tree structure

   The tree has nodes: 1 (root), 2 and 3 (children), 4 and 5 (children of 2). Total nodes = 5.

2. Step 2: Verify recursive count function output

   The function counts 1 for root plus counts of left and right subtrees, summing to 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Total nodes counted = 5 [OK]

Hint: Count all nodes including root and children [OK]

Common Mistakes:

• Counting only leaf nodes
• Missing one subtree
• Off-by-one error in count

4. Identify the error in the following TypeScript function to count total nodes in a binary tree:

function countNodes(root: TreeNode | null): number {
  if (root === null) return 1;
  return 1 + countNodes(root.left) + countNodes(root.right);
}

medium

A. Function does not count the current node

B. Base case returns 1 instead of 0 for null

C. Missing return type annotation

D. Incorrect parameter type

Solution

1. Step 1: Check base case return value

   The base case returns 1 when root is null, which incorrectly counts non-existent nodes.

2. Step 2: Understand correct base case for counting nodes

   For null nodes, the count should be 0, not 1, to avoid overcounting.

3. Final Answer:

   Base case returns 1 instead of 0 for null -> Option B

4. Quick Check:

   Base case must return 0 for null [OK]

Hint: Base case for null must return 0 nodes [OK]

Common Mistakes:

• Returning 1 for null node
• Not counting current node
• Wrong parameter or return types

5. You want to count total nodes in a binary tree but only count nodes with even values. Which modification to the standard count function is correct?

function countEvenNodes(root: TreeNode | null): number {
  if (!root) return 0;
  const leftCount = countEvenNodes(root.left);
  const rightCount = countEvenNodes(root.right);
  // What to add here?
}

hard

A. return (root.val % 2 === 0 ? 1 : 0) + leftCount + rightCount;

B. return 1 + leftCount + rightCount;

C. return leftCount + rightCount;

D. return (root.val % 2 !== 0 ? 1 : 0) + leftCount + rightCount;

Solution

1. Step 1: Check node value for evenness

   Use modulo operator to check if root.val is even; if yes, count 1, else 0.

2. Step 2: Sum counts from left, right, and current node if even

   Add leftCount, rightCount, and 1 or 0 based on current node's evenness.

3. Final Answer:

   return (root.val % 2 === 0 ? 1 : 0) + leftCount + rightCount; -> Option A

4. Quick Check:

   Count only even nodes correctly [OK]

Hint: Add 1 only if node value is even [OK]

Common Mistakes:

• Counting all nodes regardless of value
• Counting odd nodes instead
• Forgetting to add left and right counts