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

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if a graph has cycles.

DSA Go

func hasCycle(graph map[int][]int) bool {
    visited := make(map[int]bool)
    var dfs func(node int, parent int) bool
    dfs = func(node int, parent int) bool {
        visited[node] = true
        for _, neighbor := range graph[node] {
            if !visited[neighbor] {
                if dfs(neighbor, node) [1] {
                    return true
                }
            } else if neighbor != parent {
                return true
            }
        }
        return false
    }
    for node := range graph {
        if !visited[node] {
            if dfs(node, -1) {
                return true
            }
        }
    }
    return false
}

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

A== true

B!= false

C!= true

D== false

Attempts:

3 left

💡 Hint

Common Mistakes

Using '== false' instead of '== true' causes missing cycle detection.

Not checking the recursive call result leads to incorrect results.

The DFS function returns true if a cycle is found. So we check if dfs(...) == true to return true.

2fill in blank

medium

Complete the code to add an edge in an undirected graph.

DSA Go

func addEdge(graph map[int][]int, u int, v int) {
    graph[u] = append(graph[u], [1])
    graph[v] = append(graph[v], u)
}

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

Au

Bgraph[v]

Cgraph[u]

Dv

Attempts:

3 left

💡 Hint

Common Mistakes

Appending u to graph[u] instead of v.

Appending adjacency lists instead of node values.

To add an edge from u to v, append v to graph[u].

3fill in blank

hard

Fix the error in the code to check if a tree is connected.

DSA Go

func isConnected(graph map[int][]int, n int) bool {
    visited := make(map[int]bool)
    var dfs func(node int)
    dfs = func(node int) {
        visited[node] = true
        for _, neighbor := range graph[node] {
            if !visited[neighbor] {
                dfs([1])
            }
        }
    }
    dfs(0)
    return len(visited) == n
}

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

Anode

Bneighbor

Cgraph[node]

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Recursing on the current node instead of neighbor.

Using graph[node] which is a slice, not a node.

The DFS should recurse on the neighbor node, not the current node.

4fill in blank

hard

Fill both blanks to create a map of node degrees in a graph.

DSA Go

func nodeDegrees(graph map[int][]int) map[int]int {
    degrees := make(map[int]int)
    for node, neighbors := range graph {
        degrees[node] = [1](neighbors)
        for _, neighbor := range neighbors {
            if _, ok := degrees[neighbor]; !ok {
                degrees[neighbor] = [2]
            }
        }
    }
    return degrees
}

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

Alen

B0

C1

Dcap

Attempts:

3 left

💡 Hint

Common Mistakes

Using cap() instead of len() for length.

Initializing unseen neighbors with 1 instead of 0.

The degree of a node is the length of its neighbors slice. Initialize unseen neighbors with 0.

5fill in blank

hard

Fill all three blanks to create a function that returns true if a graph is a tree.

DSA Go

func isTree(graph map[int][]int, n int) bool {
    if len(graph) != n {
        return false
    }
    visited := make(map[int]bool)
    var dfs func(node int, parent int) bool
    dfs = func(node int, parent int) bool {
        visited[node] = true
        for _, neighbor := range graph[node] {
            if !visited[neighbor] {
                if dfs(neighbor, node) [1] {
                    return true
                }
            } else if neighbor != [2] {
                return true
            }
        }
        return false
    }
    if dfs(0, -1) [3] {
        return false
    }
    return len(visited) == n
}

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

A== true

Bparent

C== false

Dnode

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' false instead of '==' true for cycle detection.

Comparing neighbor to node instead of parent.

Returning true when dfs returns false.

Detect cycles by checking if dfs returns true. Check if neighbor is not the parent to detect back edges. If dfs returns false, no cycle found.

Practice

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

easy

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

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

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

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

Solution

1. Step 1: Understand the cycle property of trees and graphs

   A tree is a special type of graph that does not contain any cycles, meaning you cannot start at one node and return to it by following edges. Graphs, however, can have cycles.

2. Step 2: Analyze the options based on this property

   A tree has no cycles, but a graph can have cycles. correctly states that trees have no cycles while graphs can have cycles. Other options incorrectly describe properties of trees and graphs.

3. Final Answer:

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

4. Quick Check:

   Cycle presence = Tree no, Graph yes [OK]

Hint: Trees never loop back; graphs can have loops [OK]

Common Mistakes:

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

2. Which of the following Go code snippets correctly declares an adjacency list for an undirected tree with 3 nodes?

easy

A. adj := map[int]int{1: 2, 2: 3, 3: 1}

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

C. adj := map[int][]int{1: {2}, 2: {1, 3}, 3: {2}}

D. adj := map[int][]int{1: {2, 3}, 2: {3}, 3: {1}}

Solution

1. Step 1: Understand adjacency list representation in Go

   An adjacency list for a graph or tree is usually a map from node to a slice of connected nodes. For an undirected tree, edges are bidirectional.

2. Step 2: Check each option for correctness

   adj := map[int][]int{1: {2}, 2: {1, 3}, 3: {2}} correctly shows node 1 connected to 2, node 2 connected to 1 and 3, and node 3 connected to 2, representing an undirected tree. Options A and C use incorrect types or structures. adj := map[int][]int{1: {2, 3}, 2: {3}, 3: {1}} incorrectly connects nodes, creating extra edges.

3. Final Answer:

   adj := map[int][]int{1: {2}, 2: {1, 3}, 3: {2}} -> Option C

4. Quick Check:

   Adjacency list with bidirectional edges [OK]

Hint: Adjacency list uses map[int][]int for undirected edges [OK]

Common Mistakes:

• Using map[int]int instead of map[int][]int
• Not including both directions for undirected edges
• Using slices of slices incorrectly

3. Given the following Go code representing a graph, what will be the output of the fmt.Println(hasCycle(adj)) function if it detects cycles?

adj := map[int][]int{
  1: {2},
  2: {1, 3},
  3: {2, 4},
  4: {3, 1},
}

func hasCycle(graph map[int][]int) bool {
  visited := make(map[int]bool)
  var dfs func(int, int) bool
  dfs = func(node, parent int) bool {
    visited[node] = true
    for _, neighbor := range graph[node] {
      if !visited[neighbor] {
        if dfs(neighbor, node) {
          return true
        }
      } else if neighbor != parent {
        return true
      }
    }
    return false
  }
  for node := range graph {
    if !visited[node] {
      if dfs(node, -1) {
        return true
      }
    }
  }
  return false
}

medium

A. true

B. false

C. Compilation error

D. Panic at runtime

Solution

1. Step 1: Analyze the graph structure for cycles

   The adjacency list shows nodes 1-2-3-4 connected in a cycle: 1->2->3->4->1.

2. Step 2: Understand the cycle detection function

   The function uses DFS to detect cycles by checking if a visited neighbor is not the parent. Since the graph has a cycle, the function returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Cycle detected = true [OK]

Hint: Cycle detection returns true if back edge found [OK]

Common Mistakes:

• Ignoring the back edge condition
• Assuming no cycles in undirected graphs
• Confusing parent node check in DFS

4. The following Go code attempts to check if a given graph is a tree. It returns false even for a valid tree. What is the bug?

func isTree(graph map[int][]int) bool {
  visited := make(map[int]bool)
  var dfs func(int, int) bool
  dfs = func(node, parent int) bool {
    visited[node] = true
    for _, neighbor := range graph[node] {
      if visited[neighbor] {
        return false
      }
      if !dfs(neighbor, node) {
        return false
      }
    }
    return true
  }
  return dfs(1, -1)
}

medium

A. It uses a map instead of a slice for visited nodes.

B. It does not mark nodes as visited.

C. It does not start DFS from node 0.

D. It does not check for cycles correctly because it returns false on any visited neighbor.

Solution

1. Step 1: Understand the DFS cycle check logic

   The code returns false immediately if a neighbor is visited, but in undirected graphs, the parent node is also visited and should be ignored.

2. Step 2: Identify the missing parent check

   The code lacks a condition to skip the parent node when checking visited neighbors, causing false cycle detection.

3. Final Answer:

   It does not check for cycles correctly because it returns false on any visited neighbor. -> Option D

4. Quick Check:

   Missing parent check causes false cycle detection [OK]

Hint: Skip parent node when checking visited neighbors [OK]

Common Mistakes:

• Not ignoring parent node in DFS
• Assuming visited means cycle without context
• Starting DFS from wrong node

5. You have a graph represented by an adjacency list in Go. You want to verify if it is a tree. Which combination of checks correctly confirms this? A tree must be connected and have no cycles. Which code snippet below correctly implements this logic?

func isTree(graph map[int][]int) bool {
  visited := make(map[int]bool)
  var dfs func(int, int) bool
  dfs = func(node, parent int) bool {
    visited[node] = true
    for _, neighbor := range graph[node] {
      if !visited[neighbor] {
        if !dfs(neighbor, node) {
          return false
        }
      } else if neighbor != parent {
        return false
      }
    }
    return true
  }
  if !dfs(1, -1) {
    return false
  }
  for node := range graph {
    if !visited[node] {
      return false
    }
  }
  return true
}

hard

A. The code incorrectly assumes node 0 as root.

B. The code correctly checks for cycles and connectivity to confirm a tree.

C. The code only checks cycles but not connectivity.

D. The code only checks connectivity but not cycles.

Solution

1. Step 1: Check cycle detection in DFS

   The DFS function returns false if it finds a back edge (visited neighbor not parent), correctly detecting cycles.

2. Step 2: Check connectivity after DFS

   After DFS from node 1, the code checks if all nodes were visited, ensuring the graph is connected.

3. Step 3: Confirm both conditions for tree

   Since the code checks both no cycles and connectivity, it correctly confirms if the graph is a tree.

4. Final Answer:

   The code correctly checks for cycles and connectivity to confirm a tree. -> Option B

5. Quick Check:

   Cycle check + connectivity = tree confirmation [OK]

Hint: Check no cycles and all nodes visited for tree [OK]

Common Mistakes:

• Checking only cycles or only connectivity
• Not verifying all nodes are visited
• Assuming any connected graph is a tree