Diameter of Binary Tree in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the TreeNode class constructor.

DSA Python

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = [1]
        self.left = left
        self.right = right

Drag options to blanks, or click blank then click option'

Aright

Bleft

Cval

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning left or right to self.val instead of val.

Using a fixed number like 0 instead of the parameter val.

The constructor assigns the input value val to the node's val attribute.

2fill in blank

medium

Complete the code to calculate the height of a binary tree node recursively.

DSA Python

def height(node):
    if not node:
        return 0
    left_height = height(node.left)
    right_height = height(node.right)
    return 1 + [1](left_height, right_height)

Drag options to blanks, or click blank then click option'

Amax

Bmin

Csum

Dabs

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max causes wrong height calculation.

Using sum adds both heights which is incorrect for height.

The height of a node is 1 plus the maximum height of its left and right children.

3fill in blank

hard

Fix the error in the recursive diameter calculation by completing the return statement.

DSA Python

def diameter(node):
    if not node:
        return 0
    left_height = height(node.left)
    right_height = height(node.right)
    left_diameter = diameter(node.left)
    right_diameter = diameter(node.right)
    return max(left_height + right_height, [1](left_diameter, right_diameter))

Drag options to blanks, or click blank then click option'

Amin

Babs

Csum

Dmax

Attempts:

3 left

💡 Hint

Common Mistakes

Using min causes wrong diameter calculation.

Using sum adds diameters incorrectly.

The diameter is the maximum of the path through the root and the largest diameter in the subtrees.

4fill in blank

hard

Fill both blanks to complete the optimized diameter calculation using a helper function that returns height and diameter.

DSA Python

def diameter_helper(node):
    if not node:
        return (0, 0)  # height, diameter
    left_height, left_diameter = diameter_helper(node.left)
    right_height, right_diameter = diameter_helper(node.right)
    current_height = 1 + [1](left_height, right_height)
    current_diameter = max(left_height + right_height, [2](left_diameter, right_diameter))
    return (current_height, current_diameter)

Drag options to blanks, or click blank then click option'

Amax

Bmin

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max for height or diameter.

Mixing up height and diameter calculations.

Height is 1 plus max of left and right heights. Diameter is max of path through root and largest subtree diameter.

5fill in blank

hard

Fill all three blanks to complete the main function that returns the diameter of the binary tree.

DSA Python

def diameter_of_binary_tree(root):
    def helper(node):
        if not node:
            return (0, 0)  # height, diameter
        left_height, left_diameter = helper(node.left)
        right_height, right_diameter = helper(node.right)
        current_height = 1 + [1](left_height, right_height)
        current_diameter = max(left_height + right_height, [2](left_diameter, right_diameter))
        return (current_height, current_diameter)
    return [3](root)[1]

Drag options to blanks, or click blank then click option'

Amin

Bmax

Chelper

Ddiameter_helper

Attempts:

3 left

💡 Hint

Common Mistakes

Using min instead of max for height or diameter.

Calling a wrong function name or forgetting to return diameter.

The helper function calculates height and diameter recursively. The main function returns the diameter part of the tuple.

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