Diameter of Binary Tree in DSA Python - Build from Scratch

Diameter of Binary Tree

📖 Scenario: You are working on a network system where each node represents a server. The longest path between any two servers is important to understand the maximum delay in communication.This longest path is called the diameter of the network tree.

🎯 Goal: Build a program to find the diameter of a binary tree representing the network.The diameter is the number of edges on the longest path between any two nodes.

📋 What You'll Learn

Create a class called TreeNode to represent each node in the binary tree with attributes val, left, and right.

Create a binary tree with the exact structure given.

Create a helper function depth that calculates the depth of a node and updates the diameter.

Calculate and print the diameter of the binary tree.

💡 Why This Matters

🌍 Real World

Network engineers use tree structures to model communication paths. Finding the longest path helps identify maximum delay or bottlenecks.

💼 Career

Understanding tree traversal and diameter calculation is useful for software engineers working on network design, data structures, and algorithms.

1

Create the TreeNode class and build the binary tree

Create a class called TreeNode with an __init__ method that takes val and sets left and right to None. Then create the binary tree with root node value 1, left child 2, right child 3, left child of 2 as 4, and right child of 2 as 5.

DSA Python

# Define TreeNode class and create the binary tree as described
# Your code here

Hint

Define the TreeNode class first. Then create nodes and link them as left and right children.

2

Create a variable to store the diameter

Create a variable called diameter and set it to 0 to keep track of the longest path found so far.

DSA Python

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

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

# Create diameter variable
# Your code here

Hint

Use a simple variable to store the maximum diameter found during traversal.

3

Write the depth function to calculate diameter

Define a function called depth that takes a node as input. Inside, if node is None, return 0. Otherwise, recursively find the depth of left and right children by calling depth(node.left) and depth(node.right). Update the global diameter variable with the maximum of its current value and the sum of left and right depths. Return the maximum of left and right depths plus 1.

DSA Python

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

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

diameter = 0

def depth(node):
    # Your code here

Hint

Use recursion to find depths. Update diameter by adding left and right depths.

4

Calculate and print the diameter

Call the depth function with root as argument to start the calculation. Then print the value of diameter.

DSA Python

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

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

diameter = 0

def depth(node):
    global diameter
    if node is None:
        return 0
    left_depth = depth(node.left)
    right_depth = depth(node.right)
    diameter = max(diameter, left_depth + right_depth)
    return max(left_depth, right_depth) + 1

# Call depth with root and print diameter
# Your code here

Hint

Start the depth calculation from the root and then print the diameter variable.

Practice

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

easy

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

B. The number of nodes in the tree

C. The height of the tree

D. The number of leaf nodes

Solution

1. Step 1: Understand the diameter definition

   The diameter is the longest path between any two nodes, not just from root to leaf.

2. Step 2: Differentiate from height and node count

   Height is longest root-to-leaf path; node count is total nodes; diameter can be longer than height.

3. Final Answer:

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

4. Quick Check:

   Diameter = longest path length [OK]

Hint: Diameter is longest path between any two nodes, not just height [OK]

Common Mistakes:

• Confusing diameter with height
• Thinking diameter counts nodes instead of edges
• Assuming diameter is always root to leaf

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

easy

A. def diameter(root: int) -> TreeNode:

B. def diameter(root: Optional[TreeNode]) -> int:

C. def diameter(root: TreeNode) -> str:

D. def diameter(root: List[int]) -> int:

Solution

1. Step 1: Identify correct parameter and return types

   Function takes a tree node or None and returns an integer diameter length.

2. Step 2: Check each option's signature

   def diameter(root: Optional[TreeNode]) -> int: matches expected input (node or None) and output (int). Others mismatch types.

3. Final Answer:

   def diameter(root: Optional[TreeNode]) -> int: -> Option B

4. Quick Check:

   Input node, output int diameter [OK]

Hint: Diameter function inputs root node, returns int length [OK]

Common Mistakes:

• Using wrong input type like int or list
• Returning wrong type like str or TreeNode
• Missing Optional for nullable root

3. What is 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 diameter(root):
    diameter_val = 0
    def depth(node):
        nonlocal diameter_val
        if not node:
            return 0
        left = depth(node.left)
        right = depth(node.right)
        diameter_val = max(diameter_val, left + right)
        return max(left, right) + 1
    depth(root)
    return diameter_val

root = TreeNode(1, TreeNode(2, TreeNode(4), TreeNode(5)), TreeNode(3))
print(diameter(root))

medium

A. 2

B. 4

C. 3

D. 5

Solution

1. Step 1: Understand tree structure and depth calls

   Tree has root 1, left subtree with nodes 2,4,5 and right subtree with node 3. Longest path is 4->2->1->3 or 5->2->1->3 with length 3 edges.

2. Step 2: Calculate diameter_val during recursion

   At node 2, left and right depths are 1 each, sum 2; at root, left depth 2, right depth 1, sum 3 updates diameter_val to 3.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Longest path edges = 3 [OK]

Hint: Diameter counts edges on longest path, not nodes [OK]

Common Mistakes:

• Counting nodes instead of edges for diameter
• Ignoring nonlocal update of diameter_val
• Misunderstanding tree structure

4. Find the error in the following diameter function and select the fix:

def diameter(root):
    diameter_val = 0
    def depth(node):
        if not node:
            return 0
        left = depth(node.left)
        right = depth(node.right)
        diameter_val = max(diameter_val, left + right)
        return max(left, right) + 1
    depth(root)
    return diameter_val

medium

A. Initialize diameter_val inside depth() function

B. Change 'diameter_val' to a global variable

C. Return diameter_val inside depth() function

D. Add 'nonlocal diameter_val' inside depth() to update outer variable

Solution

1. Step 1: Identify variable scope issue

   diameter_val is assigned inside depth() but not declared nonlocal, so assignment creates local variable, not updating outer diameter_val.

2. Step 2: Fix by declaring nonlocal

   Adding 'nonlocal diameter_val' inside depth() allows updating the outer variable correctly.

3. Final Answer:

   Add 'nonlocal diameter_val' inside depth() to update outer variable -> Option D

4. Quick Check:

   Use nonlocal to modify outer variable [OK]

Hint: Use nonlocal to update outer variable inside nested function [OK]

Common Mistakes:

• Forgetting nonlocal causes no diameter update
• Using global unnecessarily
• Returning diameter_val inside depth() breaks logic

5. Given a binary tree where some nodes have only one child, how does the diameter calculation handle paths through these nodes? Choose the correct statement about the diameter in such cases.

hard

A. Diameter counts the longest path even if it passes through nodes with only one child

B. Diameter ignores nodes with only one child when calculating longest path

C. Diameter is always the height of the tree when nodes have one child

D. Diameter equals the number of leaf nodes in the tree

Solution

1. Step 1: Understand diameter path definition

   Diameter is longest path between any two nodes, regardless of node children count.

2. Step 2: Consider nodes with one child

   Paths can pass through nodes with one child; these nodes do not exclude paths from consideration.

3. Final Answer:

   Diameter counts the longest path even if it passes through nodes with only one child -> Option A

4. Quick Check:

   Diameter includes all nodes on longest path [OK]

Hint: Diameter includes all nodes on longest path, child count doesn't exclude [OK]

Common Mistakes:

• Assuming nodes with one child are ignored
• Confusing diameter with height in skewed trees
• Thinking diameter equals leaf count