Graph vs Tree Key Structural Difference in DSA Python - Interactive Comparison Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a graph has cycles.

DSA Python

def has_cycle(graph):
    visited = set()
    def dfs(node, parent):
        visited.add(node)
        for neighbor in graph[node]:
            if neighbor not in visited:
                if dfs(neighbor, node):
                    return True
            elif neighbor != [1]:
                return True
        return False
    for node in graph:
        if node not in visited:
            if dfs(node, None):
                return True
    return False

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

Aparent

Bnode

Cneighbor

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Using the current node instead of the parent node in the cycle check.

Not checking if the neighbor is different from the parent.

The cycle check returns True if a neighbor is visited and is not the parent node, indicating a cycle.

2fill in blank

medium

Complete the code to add an edge in an undirected graph represented as adjacency list.

DSA Python

def add_edge(graph, u, v):
    graph.setdefault(u, []).append([1])
    graph.setdefault(v, []).append(u)

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

Au

Bv

Cgraph

Dedge

Attempts:

3 left

💡 Hint

Common Mistakes

Appending u to u's own adjacency list.

Appending the whole graph instead of the node.

To add an edge from u to v, append v to u's adjacency list.

3fill in blank

hard

Fix the error in the code that checks if a tree is a valid tree (connected and no cycles).

DSA Python

def valid_tree(n, edges):
    graph = {i: [] for i in range(n)}
    for u, v in edges:
        graph[u].append(v)
        graph[v].append(u)
    visited = set()
    def dfs(node, parent):
        visited.add(node)
        for neighbor in graph[node]:
            if neighbor not in visited:
                if dfs(neighbor, node):
                    return True
            elif neighbor == [1]:
                return True
        return False
    if dfs(0, -1):
        return False
    return len(visited) == n

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

Anode

Bgraph

Cneighbor

Dparent

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if neighbor == node instead of neighbor != parent.

Not handling the parent node correctly.

The cycle check should return True if a neighbor is visited and is not the parent node, so the condition must check if neighbor != parent. Here, the error is that it checks neighbor == node, which is wrong.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps each node to its degree (number of edges).

DSA Python

degrees = {node: [1] for node in graph if len(graph[node]) [2] 0}

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

Alen(graph[node])

B>

C>=

Dgraph[node]

Attempts:

3 left

💡 Hint

Common Mistakes

Using the adjacency list itself instead of its length.

Using >= instead of > which includes nodes with zero degree.

The degree of a node is the length of its adjacency list. We want nodes with degree greater than 0.

5fill in blank

hard

Fill all three blanks to create a set comprehension that collects all leaf nodes (nodes with degree 1).

DSA Python

leaves = {node for node in graph if len(graph[node]) [1] 1 and node not in [2] and node != [3]

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

A==

Bvisited

Cparent

Dgraph

Attempts:

3 left

💡 Hint

Common Mistakes

Using != instead of == for degree check.

Not excluding visited or parent nodes.

Leaves have degree exactly 1. We exclude nodes already visited and the parent node.

Practice

1. Which of the following is a key structural difference between a graph and a tree?

easy

A. A graph always has a root node, but a tree does not.

B. A graph is always directed, but a tree is always undirected.

C. A tree can have multiple edges between the same nodes, but a graph cannot.

D. A tree has no cycles, but a graph can have cycles.

Solution

1. Step 1: Understand cycle presence in trees and graphs

   A tree is a special type of graph that does not contain any cycles. Graphs can have cycles or not.

2. Step 2: Analyze other options for correctness

   Graphs do not always have a root node, and trees do not have multiple edges between the same nodes. Also, graphs can be directed or undirected, and trees are usually undirected but can be represented as directed.

3. Final Answer:

   A tree has no cycles, but a graph can have cycles. -> Option D

4. Quick Check:

   Cycle presence = Tree no cycles, Graph can have cycles [OK]

Hint: Trees never have cycles; graphs can have cycles [OK]

Common Mistakes:

• Thinking graphs always have a root node
• Believing trees can have cycles
• Confusing directed and undirected properties

2. Which syntax correctly represents an undirected tree using an adjacency list in Python?

easy

A. tree = {1: [2], 2: [3], 3: [1]}

B. tree = {1: [2, 3], 2: [3], 3: [1]}

C. tree = {1: [2, 3], 2: [1], 3: [1]}

D. tree = {1: [2, 3], 2: [1, 3], 3: [2]}

Solution

1. Step 1: Understand adjacency list for undirected tree

   In an undirected tree, edges are bidirectional. So if node 1 connects to 2 and 3, nodes 2 and 3 must connect back to 1.

2. Step 2: Check each option for bidirectional edges

   tree = {1: [2, 3], 2: [1], 3: [1]} shows node 1 connected to 2 and 3, and nodes 2 and 3 connected back to 1. Others miss some connections or add extra edges.

3. Final Answer:

   tree = {1: [2, 3], 2: [1], 3: [1]} -> Option C

4. Quick Check:

   Bidirectional edges in adjacency list = tree = {1: [2, 3], 2: [1], 3: [1]} [OK]

Hint: Undirected edges appear in both nodes' lists [OK]

Common Mistakes:

• Missing reverse connections in adjacency list
• Adding extra edges that create cycles
• Confusing directed and undirected adjacency lists

3. What will be the output of the following Python code representing a graph?

graph = {1: [2], 2: [3], 3: [1]}
visited = set()
def dfs(node):
    if node in visited:
        return
    visited.add(node)
    for neighbor in graph[node]:
        dfs(neighbor)
dfs(1)
print(sorted(visited))

medium

A. []

B. [1, 2, 3]

C. [1]

D. [1, 2]

Solution

1. Step 1: Trace DFS starting from node 1

   DFS visits node 1, then its neighbor 2, then 2's neighbor 3, then 3's neighbor 1 which is already visited, so stops.

2. Step 2: Collect visited nodes and sort

   Visited nodes are {1, 2, 3}. Sorting gives [1, 2, 3].

3. Final Answer:

   [1, 2, 3] -> Option B

4. Quick Check:

   DFS visits all connected nodes = [1, 2, 3] [OK]

Hint: DFS visits all reachable nodes from start [OK]

Common Mistakes:

• Stopping DFS too early
• Not adding nodes to visited set
• Confusing order of visited nodes

4. Find the error in this Python code that tries to check if a graph is a tree:

def is_tree(graph):
    visited = set()
    def dfs(node, parent):
        visited.add(node)
        for neighbor in graph[node]:
            if neighbor != parent:
                if neighbor in visited:
                    return False
                if not dfs(neighbor, node):
                    return False
        return True
    return dfs(0, -1) and len(visited) == len(graph)

medium

A. The DFS function does not return the result of recursive calls.

B. The graph keys should start from 1, not 0.

C. The visited set is not initialized properly.

D. The parent parameter is not used correctly.

Solution

1. Step 1: Analyze DFS recursive calls

   The DFS function calls itself recursively but does not return the result of these calls, so it ignores false returns from deeper calls.

2. Step 2: Identify missing return propagation

   Without returning the recursive call's result, the function always returns true at the end, missing cycle detection.

3. Final Answer:

   The DFS function does not return the result of recursive calls. -> Option A

4. Quick Check:

   Recursive return missing causes incorrect cycle detection [OK]

Hint: Always return recursive call results in DFS [OK]

Common Mistakes:

• Ignoring return values of recursive DFS calls
• Assuming node numbering starts at 0
• Misusing parent parameter in cycle check

5. Given a graph represented as an adjacency list, which condition must be true for it to be a tree?

hard

A. It must be connected and have exactly n-1 edges, where n is the number of nodes.

B. It must have no cycles and at least n edges.

C. It must be disconnected and have fewer than n-1 edges.

D. It must have cycles and exactly n edges.

Solution

1. Step 1: Recall tree properties

   A tree with n nodes must be connected (all nodes reachable) and have exactly n-1 edges to avoid cycles.

2. Step 2: Check each option against these properties

   It must be connected and have exactly n-1 edges, where n is the number of nodes. matches these conditions. Others either allow cycles, disconnectedness, or wrong edge counts.

3. Final Answer:

   It must be connected and have exactly n-1 edges, where n is the number of nodes. -> Option A

4. Quick Check:

   Tree = connected + edges = n-1 [OK]

Hint: Tree edges = nodes - 1 and connected [OK]

Common Mistakes:

• Allowing cycles in a tree
• Confusing edge count with nodes
• Ignoring connectivity requirement