Diameter of Binary Tree in DSA Python - Time & Space Complexity

Time Complexity: Diameter of Binary Tree

O(n)

Understanding Time Complexity

We want to understand how the time needed to find the diameter of a binary tree grows as the tree gets bigger.

The question is: how does the work increase when the number of nodes increases?

Scenario Under Consideration

Analyze the time complexity 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_of_binary_tree(root):
    diameter = 0
    def depth(node):
        nonlocal diameter
        if not node:
            return 0
        left = depth(node.left)
        right = depth(node.right)
        diameter = max(diameter, left + right)
        return 1 + max(left, right)
    depth(root)
    return diameter
    

This code finds the longest path between any two nodes in a binary tree by calculating depths and updating the diameter.

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 exactly one time.

How Execution Grows With Input

As the number of nodes grows, the function visits each node 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 work grows linearly as the tree size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the diameter grows in direct proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "The diameter calculation takes more than linear time because it looks at paths between nodes."

[OK] Correct: The code visits each node once and calculates depths on the way back, so it does not repeatedly check paths, keeping the time linear.

Interview Connect

Understanding this helps you explain how recursion can efficiently solve tree problems by visiting nodes once, a key skill in many coding challenges.

Self-Check

"What if we changed the code to calculate diameter by checking all pairs of nodes separately? How would the time complexity change?"

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