Count Total Nodes in Binary Tree in DSA Javascript - Execution Trace

Concept Flow - Count Total Nodes in Binary Tree

Start at root node

↓

Is node null?

Yes→Return 0

No↓

Count left subtree nodes

↓

Count right subtree nodes

↓

Add 1 (current node) + left count + right count

↓

Return total count

The process starts at the root, checks if the node is null, then recursively counts nodes in left and right subtrees, adding 1 for the current node.

Execution Sample

DSA Javascript

function countNodes(node) {
  if (node === null) return 0;
  return 1 + countNodes(node.left) + countNodes(node.right);
}

This function counts all nodes in a binary tree by recursively summing nodes from left and right children plus the current node.

Execution Table

Step	Operation	Node Visited	Left Count	Right Count	Total Count	Tree State
1	Visit root	A	-	-	-	A / \ B C
2	Visit left child	B	-	-	-	A / \ B C
3	Visit B's left child	D	-	-	-	A / \ B C / D
4	D is leaf, left null	null	0	-	-	A / \ B C / D
5	D is leaf, right null	null	-	0	-	A / \ B C / D
6	Count D nodes	D	0	0	1	A / \ B C / D
7	Visit B's right child	null	0	-	-	A / \ B C / D
8	Count B nodes	B	1	0	2	A / \ B C / D
9	Visit right child	C	-	-	-	A / \ B C / D
10	Visit C's left child	null	0	-	-	A / \ B C / D
11	Visit C's right child	E	-	-	-	A / \ B C / D \ E
12	E is leaf, left null	null	0	-	-	A / \ B C / D \ E
13	E is leaf, right null	null	-	0	-	A / \ B C / D \ E
14	Count E nodes	E	0	0	1	A / \ B C / D \ E
15	Count C nodes	C	0	1	2	A / \ B C / D \ E
16	Count root nodes	A	2	2	5	A / \ B C / D \ E
17	Return total count	A	2	2	5	Final count: 5 nodes

💡 All nodes visited; recursion ends when null nodes are reached.

Variable Tracker

Variable	Start	After Step 6	After Step 8	After Step 14	After Step 15	After Step 16	Final
node	A (root)	D (leaf)	B (left child)	E (leaf)	C (right child)	A (root)	null (end)
leftCount	-	0	1	0	0	2	2
rightCount	-	0	0	0	1	2	2
totalCount	-	1	2	1	2	5	5

Key Moments - 3 Insights

Why do we return 0 when the node is null?

Returning 0 at null nodes (see Step 4 and 5) stops recursion and counts no nodes beyond leaves.

How does the function add counts from left and right subtrees?

At each node (e.g., Step 16), it adds 1 (current node) plus leftCount and rightCount from recursive calls.

Why does the total count at root equal 5?

Because it sums all nodes counted in left subtree (2), right subtree (2), plus 1 for root itself (Step 16).

Visual Quiz - 3 Questions

Look at the execution table, what is the totalCount at Step 8?

A 1

B 2

C 0

D 3

💡 Hint

Check the 'totalCount' column at Step 8 in the execution_table.

At which step does the function visit node E?

A Step 14

B Step 10

C Step 11

D Step 15

💡 Hint

Look for 'Visit C's right child' operation in the execution_table.

If node D had a left child, how would the totalCount at Step 6 change?

A It would increase

B It would decrease

C It would remain 1

D It would become 0

💡 Hint

Adding a child node increases the count returned at that subtree (see how counts add in execution_table).

Concept Snapshot

Count Total Nodes in Binary Tree:
- Use recursion starting at root
- If node is null, return 0
- Else return 1 + count(left) + count(right)
- Counts all nodes including root
- Stops at leaves with null children

Full Transcript

To count total nodes in a binary tree, start at the root node. If the node is null, return zero because there are no nodes to count. Otherwise, count nodes in the left subtree recursively, then count nodes in the right subtree recursively. Add these counts together and add one for the current node. This process repeats until all nodes are counted. The final result is the total number of nodes in the tree.

Practice

1. What does the function to count total nodes in a binary tree do?

easy

A. It returns the total number of nodes in the tree.

B. It returns the height of the tree.

C. It returns the sum of all node values.

D. It returns the number of leaf nodes only.

Solution

1. Step 1: Understand the function's purpose

   The function is designed to count every node in the binary tree, including root, internal, and leaf nodes.

2. Step 2: Differentiate from other tree properties

   Counting nodes is different from calculating height, sum of values, or counting leaves, which are other tree properties.

3. Final Answer:

   It returns the total number of nodes in the tree. -> Option A

4. Quick Check:

   Counting nodes = total nodes [OK]

Hint: Count all nodes, not just leaves or height [OK]

Common Mistakes:

• Confusing node count with tree height
• Counting only leaf nodes
• Summing node values instead of counting

2. Which of the following is the correct base case syntax in JavaScript for counting nodes in a binary tree recursively?

easy

A. if (node == undefined) return 1;

B. if (node === null) return 0;

C. if (node === undefined) return null;

D. if (node === null) return null;

Solution

1. Step 1: Identify the base case for recursion

   When the node is null (no node), the count should be zero because there is no node to count.

2. Step 2: Check correct JavaScript syntax

   Using strict equality (===) to check for null and returning 0 is the correct approach.

3. Final Answer:

   if (node === null) return 0; -> Option B

4. Quick Check:

   Base case null node returns 0 [OK]

Hint: Base case returns 0 for null nodes [OK]

Common Mistakes:

• Returning 1 for null nodes
• Using == instead of ===
• Returning null instead of 0

3. What is the output of the following code snippet counting nodes in a binary tree?

class Node {
  constructor(val) {
    this.val = val;
    this.left = null;
    this.right = null;
  }
}

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

function countNodes(node) {
  if (node === null) return 0;
  return 1 + countNodes(node.left) + countNodes(node.right);
}

console.log(countNodes(root));

medium

A. 2

B. 3

C. 5

D. 4

Solution

1. Step 1: Count nodes in the tree structure

   The tree has nodes: 1 (root), 2 (left child), 3 (right child), and 4 (left child of 2). Total nodes = 4.

2. Step 2: Apply recursive countNodes function

   Function counts 1 for current node plus counts of left and right subtrees recursively, resulting in 4.

3. Final Answer:

   4 -> Option D

4. Quick Check:

   Nodes counted = 4 [OK]

Hint: Add 1 + left subtree + right subtree counts [OK]

Common Mistakes:

• Forgetting to count the root node
• Miscounting child nodes
• Confusing left and right subtree counts

4. Find the error in this code snippet that tries to count total nodes in a binary tree:

function countNodes(node) {
  if (node == null) return 1;
  return 1 + countNodes(node.left) + countNodes(node.right);
}

medium

A. Base case returns 1 instead of 0 for null node.

B. Missing addition of 1 for current node in return statement.

C. Using == instead of === causes error.

D. Function should return node.val instead of count.

Solution

1. Step 1: Check base case return value

   Base case returns 1 when node is null, which incorrectly counts non-existent nodes.

2. Step 2: Understand correct base case

   Base case should return 0 for null nodes because they do not exist and should not add to count.

3. Final Answer:

   Base case returns 1 instead of 0 for null node. -> Option A

4. Quick Check:

   Null node count must be 0 [OK]

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

Common Mistakes:

• Returning 1 for null nodes
• Not adding 1 for current node
• Confusing equality operators

5. Given a binary tree where some nodes have only one child, how would you modify the countNodes function to count all nodes correctly even if some children are missing?

function countNodes(node) {
  if (node === null) return 0;
  return 1 + countNodes(node.left) + countNodes(node.right);
}

hard

A. Add a global counter variable outside the function.

B. Add checks to count only nodes with two children.

C. No change needed; function already counts all nodes correctly.

D. Change base case to return 1 for null nodes.

Solution

1. Step 1: Analyze current function behavior

   The function returns 0 for null nodes and adds 1 for each existing node plus counts of left and right children recursively.

2. Step 2: Confirm it handles missing children

   Since null children return 0, missing children do not add to count, so all existing nodes are counted correctly.

3. Final Answer:

   No change needed; function already counts all nodes correctly. -> Option C

4. Quick Check:

   Function counts all nodes including those with missing children [OK]

Hint: Null children return 0, so all nodes count correctly [OK]

Common Mistakes:

• Thinking missing children need special counting
• Changing base case incorrectly
• Using global counters unnecessarily