Tree vs Array vs Linked List When Hierarchy Matters in DSA Python - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a list representing a simple hierarchy of fruits.

DSA Python

fruits = ['Apple', [1], 'Banana']

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

A'Tomato'

B'Orange'

C'Potato'

D'Carrot'

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing a vegetable instead of a fruit.

The list represents fruits, so 'Orange' fits the hierarchy correctly.

2fill in blank

medium

Complete the code to link nodes in a singly linked list representing a hierarchy.

DSA Python

class Node:
    def __init__(self, value):
        self.value = value
        self.next = [1]

node1 = Node('Root')
node2 = Node('Child')
node1.next = node2

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

Anode2

Bnode1

Cself

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to self or another node prematurely.

Initially, the next pointer should be None until linked to another node.

3fill in blank

hard

Fix the error in the tree node initialization to correctly set children as an empty list.

DSA Python

class TreeNode:
    def __init__(self, value):
        self.value = value
        self.children = [1]

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

ANone

B{}

C[]

D()

Attempts:

3 left

💡 Hint

Common Mistakes

Using None or other data types that don't support multiple children.

Children should be an empty list to hold multiple child nodes.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps node values to their number of children.

DSA Python

child_counts = {node.value: [1] for node in nodes if len(node.children) [2] 0}

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

Alen(node.children)

Bnode.children

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using the children list itself instead of its length.

Using equality instead of greater than.

We want the count of children for nodes that have more than zero children.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps uppercase node values to their first child value if it exists.

DSA Python

result = {node.value[1]: node.children[2].value for node in nodes if node.children and [3](node.children) > 0}

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

A.upper()

B[0]

Clen()

D.pop()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .pop() which removes elements.

Not checking children count properly.

We convert node values to uppercase, access the first child with [0], and check children count with len().

Practice

1. Which data structure is best suited to represent hierarchical relationships like a family tree?

easy

A. Linked List

B. Array

C. Stack

D. Tree

Solution

1. Step 1: Understand hierarchy representation

   Hierarchies have parent-child relationships, which are naturally modeled by trees.

2. Step 2: Compare data structures

   Arrays and linked lists are linear and do not represent branching well, unlike trees.

3. Final Answer:

   Tree -> Option D

4. Quick Check:

   Hierarchy = Tree [OK]

Hint: Hierarchy means branching; trees model branching best [OK]

Common Mistakes:

• Choosing array for hierarchy
• Confusing linked list with tree
• Ignoring parent-child relationships

2. Which of the following is the correct way to declare a linked list node in Python?

easy

A. class Node:
def __init__(self, data):
self.data = data
self.prev = None

B. class Node:
def __init__(self, data):
self.data = data
self.next = None

C. class Node:
def __init__(self, data):
self.next = data
self.data = None

D. class Node:
def __init__(self):
self.data = None
self.next = None

Solution

1. Step 1: Identify linked list node structure

   A linked list node stores data and a pointer to the next node.

2. Step 2: Check each option

   class Node: def __init__(self, data): self.data = data self.next = None correctly assigns data and next pointer; others have incorrect attributes or missing data.

3. Final Answer:

   class Node: def __init__(self, data): self.data = data self.next = None -> Option B

4. Quick Check:

   Node has data and next pointer [OK]

Hint: Node must have data and next pointer for linked list [OK]

Common Mistakes:

• Swapping data and next attributes
• Missing next pointer
• Using prev pointer for singly linked list

3. You need to store a company's organizational chart with multiple levels of management and employees. Which data structure is best and why? A) Array because it stores data contiguously B) Linked List because it allows easy insertion C) Stack because it follows last-in-first-out order D) Tree because it naturally models multiple levels and branches

easy

A. Tree because it naturally models multiple levels and branches

B. Linked List because it allows easy insertion

C. Stack because it follows last-in-first-out order

D. Array because it stores data contiguously

Solution

1. Step 1: Understand organizational chart structure

   It has multiple levels with managers and subordinates, forming a hierarchy.

2. Step 2: Match data structure to hierarchy

   Tree structure models multiple levels and branches naturally, unlike arrays or linked lists which are linear.

3. Step 3: Eliminate other options

   Array is linear and fixed size; linked list is linear; stack is LIFO, not hierarchical.

4. Final Answer:

   Tree because it naturally models multiple levels and branches -> Option A

5. Quick Check:

   Hierarchy needs tree structure [OK]

Hint: Hierarchy with branches = use tree [OK]

Common Mistakes:

• Choosing array for hierarchical data
• Confusing linked list insertion with hierarchy
• Using stack for hierarchical representation

4. What will be the output of the following code that builds a simple tree and prints the root's first child's data?

class Node:
    def __init__(self, data):
        self.data = data
        self.children = []

root = Node('root')
child1 = Node('child1')
child2 = Node('child2')
root.children.append(child1)
root.children.append(child2)
print(root.children[0].data)

medium

A. child1

B. root

C. child2

D. Error

Solution

1. Step 1: Understand tree node structure

   Each node has data and a list of children nodes.

2. Step 2: Trace code execution

   Root has two children: child1 and child2. root.children[0] is child1, so printing its data outputs 'child1'.

3. Final Answer:

   child1 -> Option A

4. Quick Check:

   First child data = child1 [OK]

Hint: Index 0 child of root is first added child [OK]

Common Mistakes:

• Printing root data instead of child
• Confusing child1 and child2 order
• Assuming error due to children list

5. Find the error in this linked list insertion code:

class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

def insert_at_head(head, data):
    new_node = Node(data)
    new_node.next = head
    head = new_node
    return head

head = None
head = insert_at_head(head, 10)
head = insert_at_head(head, 20)
print(head.data)

medium

A. head is not updated outside function

B. new_node.next should be None

C. No error; prints 20

D. Node class missing prev pointer

Solution

1. Step 1: Analyze insert_at_head function

   Function creates new node, points it to current head, updates head to new node, and returns it.

2. Step 2: Check usage and output

   Head is updated by assignment after each insertion. Printing head.data after two insertions prints 20, the last inserted value.

3. Final Answer:

   No error; prints 20 -> Option C

4. Quick Check:

   Head updated correctly, output 20 [OK]

Hint: Function returns new head; assign it back [OK]

Common Mistakes:

• Assuming head updates automatically without assignment
• Confusing next pointer assignment
• Expecting error due to missing prev pointer