Why Shortest Path Is a Graph Problem Not a Tree Problem in DSA C++ - See It Work

Why Shortest Path Is a Graph Problem Not a Tree Problem

📖 Scenario: Imagine you are planning a trip between cities connected by roads. Some roads connect cities in a way that forms loops, and some cities have multiple paths to reach them. You want to find the shortest route from your home city to a destination city.This situation is like a graph, where cities are points (nodes) and roads are connections (edges). Unlike a tree, which has no loops and only one path between any two points, a graph can have many paths and loops.

🎯 Goal: You will build a simple graph with cities and roads, then find the shortest path between two cities. This will show why shortest path problems need graphs, not trees.

📋 What You'll Learn

Create a graph using an adjacency list with specific cities and roads

Add a variable to store the starting city

Implement a simple shortest path search using breadth-first search (BFS)

Print the shortest path distance from the start city to a target city

💡 Why This Matters

🌍 Real World

Finding shortest routes in maps, social networks, or communication networks often involves graphs with loops and multiple paths.

💼 Career

Understanding graph algorithms like shortest path is essential for software engineers working on navigation apps, network routing, and data analysis.

1

Create the graph with cities and roads

Create an adjacency list called graph as a std::unordered_map<std::string, std::vector<std::string>> with these exact entries: "A" connected to "B" and "C", "B" connected to "A", "C" connected to "A" and "D", and "D" connected to "C" and "B".

DSA C++

#include 
#include 
#include 
#include 

int main() {
    // Create the graph adjacency list here
    // Your code here

Hint

Use std::unordered_map<std::string, std::vector<std::string>> to store the graph. Each city key maps to a vector of connected cities.

2

Add the starting city variable

Add a variable called start of type std::string and set it to "A".

DSA C++

#include 
#include 
#include 
#include 

int main() {
    std::unordered_map> graph = {
        {"A", {"B", "C"}},
        {"B", {"A"}},
        {"C", {"A", "D"}},
        {"D", {"C", "B"}}
    };
    // Add the start city variable here
    // Your code here

Hint

Use std::string start = "A"; to set the starting city.

3

Implement BFS to find shortest path distances

Add code to perform a breadth-first search (BFS) starting from start to find the shortest distance to each city. Use a std::unordered_map<std::string, int> called distance to store distances. Initialize distance[start] to 0. Use a std::queue<std::string> called q to process cities. For each neighbor, if it is not in distance, set its distance to current city distance + 1 and push it to q.

DSA C++

#include 
#include 
#include 
#include 
#include 

int main() {
    std::unordered_map> graph = {
        {"A", {"B", "C"}},
        {"B", {"A"}},
        {"C", {"A", "D"}},
        {"D", {"C", "B"}}
    };
    std::string start = "A";
    // Implement BFS to find shortest distances
    // Your code here

Hint

Use BFS with a queue to explore neighbors level by level. Track distances in a map.

4

Print the shortest distance to city D

Print the shortest distance from start to city "D" using std::cout. The output should be exactly: Distance from A to D is 2.

DSA C++

#include 
#include 
#include 
#include 
#include 

int main() {
    std::unordered_map> graph = {
        {"A", {"B", "C"}},
        {"B", {"A"}},
        {"C", {"A", "D"}},
        {"D", {"C", "B"}}
    };
    std::string start = "A";

    std::unordered_map distance;
    std::queue q;

    distance[start] = 0;
    q.push(start);

    while (!q.empty()) {
        std::string current = q.front();
        q.pop();

        for (const std::string& neighbor : graph[current]) {
            if (distance.find(neighbor) == distance.end()) {
                distance[neighbor] = distance[current] + 1;
                q.push(neighbor);
            }
        }
    }
    // Print the shortest distance to city D
    // Your code here

Hint

Use std::cout to print the distance stored in distance["D"].

Practice

1. Why is finding the shortest path usually considered a graph problem rather than a tree problem?

easy

A. Because shortest path algorithms only work on trees.

B. Because trees always have weighted edges, graphs do not.

C. Because trees do not have nodes connected by edges.

D. Because graphs can have cycles and multiple paths between nodes, unlike trees.

Solution

1. Step 1: Understand the structure of trees and graphs

   Trees are a special type of graph with no cycles and exactly one path between any two nodes.

2. Step 2: Analyze shortest path requirements

   Shortest path problems require handling multiple paths and cycles, which only general graphs can have.

3. Final Answer:

   Because graphs can have cycles and multiple paths between nodes, unlike trees. -> Option D

4. Quick Check:

   Graphs allow multiple paths = A [OK]

Hint: Shortest path needs multiple paths, trees have only one [OK]

Common Mistakes:

• Thinking trees can have cycles
• Assuming shortest path only applies to trees
• Confusing edge weights with graph structure

2. Which of the following is the correct way to represent a weighted graph in C++ for shortest path algorithms?

easy

A. vector>> graph; // stores node and weight pairs

B. vector graph; // only stores edges without weights

C. int graph[][]; // fixed size 2D array without weights

D. list<int> graph; // stores nodes only

Solution

1. Step 1: Identify data structure for weighted edges

   Weighted graphs need to store both the connected node and the edge weight, so pairs are used.

2. Step 2: Match with C++ syntax

   vector>> allows storing adjacency lists with (node, weight) pairs.

3. Final Answer:

   vector>> graph; // stores node and weight pairs -> Option A

4. Quick Check:

   Weighted edges need pairs = A [OK]

Hint: Use pairs to store node and weight in adjacency list [OK]

Common Mistakes:

• Using vector loses weight info
• Using fixed size arrays limits graph size
• Using list<int> cannot store edges properly

3. Consider the following C++ code snippet for a graph with edges and weights:

vector>> graph(3);
graph[0].push_back({1, 4});
graph[0].push_back({2, 1});
graph[1].push_back({2, 2});
graph[2].push_back({1, 1});

// What is the weight of the edge from node 0 to node 2?

medium

A. 4

B. 2

C. 1

D. No edge exists

Solution

1. Step 1: Read edges from node 0

   graph[0] has two edges: to node 1 with weight 4, and to node 2 with weight 1.

2. Step 2: Identify weight for edge 0 to 2

   The edge from node 0 to node 2 has weight 1 as per graph[0].push_back({2, 1}).

3. Final Answer:

   1 -> Option C

4. Quick Check:

   Edge 0->2 weight = 1 [OK]

Hint: Check adjacency list entries for node 0 [OK]

Common Mistakes:

• Confusing edge weights between nodes
• Assuming no edge exists from 0 to 2
• Mixing up node indices

4. The following code attempts to find the shortest path in a graph but has a bug:

vector>> graph(3);
graph[0].push_back({1, 5});
graph[1].push_back({2, 3});

int dist[3] = {INT_MAX, INT_MAX, INT_MAX};
dist[0] = 0;

for (int i = 0; i < 3; i++) {
  for (auto edge : graph[i]) {
    int v = edge.first;
    int w = edge.second;
    if (dist[i] + w < dist[v]) {
      dist[v] = dist[i] + w;
    }
  }
}

What is the main issue with this code?

medium

A. It does not initialize distances correctly.

B. It does not relax edges multiple times to find shortest paths.

C. It uses wrong data structure for graph.

D. It updates distances without checking for negative weights.

Solution

1. Step 1: Understand the relaxation process

   Shortest path algorithms like Bellman-Ford require multiple passes to relax edges and update distances.

2. Step 2: Analyze the loop structure

   The code only relaxes edges once in a single pass, which may not find shortest paths if multiple updates are needed.

3. Final Answer:

   It does not relax edges multiple times to find shortest paths. -> Option B

4. Quick Check:

   Single pass relaxation misses updates = D [OK]

Hint: Relax edges multiple times for shortest path [OK]

Common Mistakes:

• Assuming one pass is enough
• Ignoring need for repeated relaxation
• Confusing initialization with relaxation

5. You have a graph with cycles and multiple paths between nodes. Why can't you use a tree-based shortest path algorithm here, and what approach should you use instead?

hard

A. Because trees have no cycles, use Dijkstra's algorithm on the graph instead.

B. Because trees have weighted edges, use BFS on the graph instead.

C. Because trees have cycles, use DFS on the graph instead.

D. Because trees are disconnected, use Prim's algorithm on the graph instead.

Solution

1. Step 1: Understand tree vs graph properties

   Trees have no cycles and exactly one path between nodes, so tree algorithms assume no alternative paths.

2. Step 2: Identify suitable shortest path algorithm for graphs

   Graphs with cycles and multiple paths require algorithms like Dijkstra's that handle weighted edges and cycles correctly.

3. Final Answer:

   Because trees have no cycles, use Dijkstra's algorithm on the graph instead. -> Option A

4. Quick Check:

   Cycles need graph algorithms like Dijkstra = B [OK]

Hint: Use Dijkstra for graphs with cycles, not tree algorithms [OK]

Common Mistakes:

• Using BFS for weighted graphs
• Confusing tree cycles with graph cycles
• Choosing Prim's which is for MST, not shortest path