Diameter of Binary Tree in DSA Javascript - Build from Scratch

Diameter of Binary Tree

📖 Scenario: You are working on a program that analyzes a family tree structure. Each person is represented as a node in a binary tree, where the left and right children represent descendants. You want to find the longest path between any two people in this family tree, which is called the diameter of the tree.

🎯 Goal: Build a JavaScript program that calculates the diameter of a binary tree. The diameter is the number of nodes on the longest path between any two nodes in the tree.

📋 What You'll Learn

Create a binary tree using JavaScript objects with val, left, and right properties.

Use a helper variable to track the maximum diameter found during traversal.

Write a recursive function to calculate the height of each subtree and update the diameter.

Print the final diameter value.

💡 Why This Matters

🌍 Real World

Finding the diameter of a tree structure helps in network design, family tree analysis, and understanding the longest communication path in hierarchical data.

💼 Career

This concept is useful for software engineers working with tree data structures, such as in databases, file systems, and network routing algorithms.

1

Create the binary tree structure

Create a binary tree by defining a variable root that represents the root node. The root node should have val 1, a left child with val 2, and a right child with val 3. The left child of node 2 should have val 4, and the right child of node 2 should have val 5.

DSA Javascript

// Create the binary tree with root, left, and right nodes
// Your code here

Hint

Use nested objects to represent each node with val, left, and right properties.

2

Add a variable to track the diameter

Add a variable called maxDiameter and set it to 0. This will keep track of the longest path found so far.

DSA Javascript

const root = {
  val: 1,
  left: {
    val: 2,
    left: { val: 4, left: null, right: null },
    right: { val: 5, left: null, right: null }
  },
  right: { val: 3, left: null, right: null }
};

// Create a variable maxDiameter and set it to 0
// Your code here

Hint

This variable will be updated later when you calculate the diameter.

3

Write a recursive function to calculate height and update diameter

Write a function called height that takes a node node as input. If node is null, return 0. Otherwise, recursively find the height of the left subtree and the right subtree. Update maxDiameter with the sum of left and right heights if it is larger than the current maxDiameter. Return the height of the current node as 1 plus the maximum of left and right heights.

DSA Javascript

const root = {
  val: 1,
  left: {
    val: 2,
    left: { val: 4, left: null, right: null },
    right: { val: 5, left: null, right: null }
  },
  right: { val: 3, left: null, right: null }
};

let maxDiameter = 0;

// Write the height function below
// Your code here

Hint

Use recursion to find the height of left and right children, then update maxDiameter.

4

Calculate and print the diameter

Call the height function with root as the argument to start the calculation. Then print the value of maxDiameter using console.log(maxDiameter).

DSA Javascript

const root = {
  val: 1,
  left: {
    val: 2,
    left: { val: 4, left: null, right: null },
    right: { val: 5, left: null, right: null }
  },
  right: { val: 3, left: null, right: null }
};

let maxDiameter = 0;

function height(node) {
  if (node === null) {
    return 0;
  }
  const leftHeight = height(node.left);
  const rightHeight = height(node.right);
  maxDiameter = Math.max(maxDiameter, leftHeight + rightHeight);
  return 1 + Math.max(leftHeight, rightHeight);
}

// Call height with root and print maxDiameter
// Your code here

Hint

Calling height(root) starts the recursive calculation. The diameter is printed last.

Practice

1. What does the diameter of a binary tree represent?

easy

A. The longest path between any two nodes in the tree

B. The height of the tree

C. The number of leaf nodes

D. The total number of nodes in the tree

Solution

1. Step 1: Understand the diameter concept

   The diameter is the longest path between any two nodes, which may or may not pass through the root.

2. Step 2: Differentiate from height and node count

   Height is the longest path from root to leaf, and node count is total nodes, both different from diameter.

3. Final Answer:

   The longest path between any two nodes in the tree -> Option A

4. Quick Check:

   Diameter = Longest path between nodes [OK]

Hint: Diameter is longest path, not just height or node count [OK]

Common Mistakes:

• Confusing diameter with tree height
• Thinking diameter counts nodes instead of edges
• Assuming diameter must pass through root

2. Which of the following JavaScript function signatures is correct for computing the diameter of a binary tree node?

easy

A. function diameterTree() {}

B. function diameter(root, tree) {}

C. function binaryTreeDiameter(node, root) {}

D. function diameterOfBinaryTree(root) {}

Solution

1. Step 1: Check common parameter naming

   The function usually takes the root node as input, so a single parameter named root is standard.

2. Step 2: Validate function signature correctness

   function diameterOfBinaryTree(root) {} uses a clear, single parameter 'root' and a descriptive function name, matching common practice.

3. Final Answer:

   function diameterOfBinaryTree(root) {} -> Option D

4. Quick Check:

   Correct function signature uses root parameter [OK]

Hint: Function usually takes root node as single parameter [OK]

Common Mistakes:

• Using multiple unrelated parameters
• Incorrect or unclear function names
• Missing parameters for the tree root

3. Given the following binary tree structure and code, what is the diameter of the tree?

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

const root = new TreeNode(1,
  new TreeNode(2,
    new TreeNode(4),
    new TreeNode(5)
  ),
  new TreeNode(3)
);

function diameterOfBinaryTree(root) {
  let diameter = 0;
  function depth(node) {
    if (!node) return 0;
    const left = depth(node.left);
    const right = depth(node.right);
    diameter = Math.max(diameter, left + right);
    return Math.max(left, right) + 1;
  }
  depth(root);
  return diameter;
}

console.log(diameterOfBinaryTree(root));

medium

A. 4

B. 3

C. 2

D. 5

Solution

1. Step 1: Analyze the tree structure

   The tree has root 1, left subtree with nodes 2,4,5 and right subtree with node 3. The longest path is from node 4 to node 3 through 2 and 1 (4->2->1->3), which has length 3 edges.

2. Step 2: Trace the code logic

   The code calculates diameter as max sum of left and right depths at each node. At node 2, left and right depths are 1 each, so diameter updates to 2. At root, left depth is 2, right depth is 1, so diameter updates to 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   Longest path edges = 3 [OK]

Hint: Diameter = max sum of left and right subtree depths [OK]

Common Mistakes:

• Counting nodes instead of edges for diameter
• Ignoring paths not passing through root
• Misreading tree structure

4. The following code attempts to compute the diameter of a binary tree but returns incorrect results. What is the bug?

function diameterOfBinaryTree(root) {
  let diameter = 0;
  function depth(node) {
    if (!node) return 0;
    const left = depth(node.left);
    const right = depth(node.right);
    diameter = Math.max(diameter, left * right);
    return Math.max(left, right) + 1;
  }
  depth(root);
  return diameter;
}

medium

A. Using multiplication instead of addition for diameter update

B. Returning max depth instead of sum of depths

C. Not handling null nodes correctly

D. Missing base case for leaf nodes

Solution

1. Step 1: Identify diameter update logic

   The code updates diameter using left * right, which multiplies depths instead of adding them.

2. Step 2: Correct diameter calculation

   Diameter should be updated with left + right, representing the longest path through the node.

3. Final Answer:

   Using multiplication instead of addition for diameter update -> Option A

4. Quick Check:

   Diameter update uses addition, not multiplication [OK]

Hint: Diameter update = left depth + right depth, not multiplication [OK]

Common Mistakes:

• Multiplying depths instead of adding
• Confusing depth return with diameter update
• Ignoring base case for null nodes

5. You want to find the diameter of a binary tree but also return the path nodes that form this diameter. Which approach correctly extends the diameter calculation to return the path?

hard

A. Use breadth-first search to find the longest path from root to any leaf

B. Only compute the height of the tree and return the nodes at the deepest level

C. Track left and right subtree depths and their paths; update diameter and store combined path when a longer diameter is found

D. Count total nodes and return all leaf nodes as the diameter path

Solution

1. Step 1: Understand diameter path requirements

   Diameter path is the longest path between any two nodes, so we must track paths from left and right subtrees.

2. Step 2: Extend depth function to track paths

   Along with depths, store paths for left and right subtrees. When diameter updates, combine these paths to form the diameter path.

3. Step 3: Reject incorrect approaches

   Height alone or BFS from root won't find longest path between any two nodes. Counting leaf nodes is unrelated.

4. Final Answer:

   Track left and right subtree depths and their paths; update diameter and store combined path when a longer diameter is found -> Option C

5. Quick Check:

   Track depths and paths to get diameter path [OK]

Hint: Store paths with depths; update when diameter increases [OK]

Common Mistakes:

• Only tracking height or BFS from root
• Ignoring path storage during recursion
• Assuming diameter path is always root to leaf