C++ 宽度优先搜索代码中需要澄清，因为输出不'；这与我的理解不符

我从techiedelight找到了这段代码[https://www.techiedelight.com/breadth-first-search/]对于BFS

代码如下：

#include <iostream>
#include <queue>
#include <vector>
using namespace std;

// Data structure to store graph edges
struct Edge {
    int src, dest;
};

// Class to represent a graph object
class Graph
{
public:
    // construct a vector of vectors to represent an adjacency list
    vector<vector<int>> adjList;

    // Graph Constructor
    Graph(vector<Edge> const &edges, int N)
    {
        // resize the vector to N elements of type vector<int>
        adjList.resize(N);

        // add edges to the undirected graph
        for (auto &edge: edges)
        {
            adjList[edge.src].push_back(edge.dest);
            adjList[edge.dest].push_back(edge.src);
        }
    }
};

// Perform BFS recursively on graph
void recursiveBFS(Graph const &graph, queue<int> &q,
                    vector<bool> &discovered)
{
    if (q.empty())
        return;

    // pop front node from queue and print it
    int v = q.front();
    q.pop();
    cout << v << " ";

    // do for every edge (v -> u)
    for (int u : graph.adjList[v])
    {
        if (!discovered[u])
        {
            // mark it discovered and push it into queue
            discovered[u] = true;
            q.push(u);
        }
    }

    recursiveBFS(graph, q, discovered);
}

// Recursive C++ implementation of Breadth first search
int main()
{
    // vector of graph edges as per above diagram
    vector<Edge> edges = {
        {1, 2}, {1, 3}, {1, 4}, {2, 5}, {2, 6}, {5, 9},
        {5, 10}, {4, 7}, {4, 8}, {7, 11}, {7, 12}
        // vertex 0, 13 and 14 are single nodes
    };

    // Number of nodes in the graph
    int N = 15;

    // create a graph from edges
    Graph graph(edges, N);

    // stores vertex is discovered or not
    vector<bool> discovered(N, false);

    // create a queue used to do BFS
    queue<int> q;

    // Do BFS traversal from all undiscovered nodes to
    // cover all unconnected components of graph
    for (int i = 0; i < N; i++) {
        if (discovered[i] == false)
        {
            // mark source vertex as discovered
            discovered[i] = true;

            // push source vertex into the queue
            q.push(i);

            // start BFS traversal from vertex i
            recursiveBFS(graph, q, discovered);
        }
    }

    return 0;
}

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 

1 2 3 4 5 7 6 9 10 11 13

然而，我不明白如果顶点0、13和14不在图形中，它是如何进入输出的？此外，根据我的理解，该输入的BFS输出与我的理解不同，应如下所示：

#include <iostream>
#include <queue>
#include <vector>
using namespace std;

// Data structure to store graph edges
struct Edge {
    int src, dest;
};

// Class to represent a graph object
class Graph
{
public:
    // construct a vector of vectors to represent an adjacency list
    vector<vector<int>> adjList;

    // Graph Constructor
    Graph(vector<Edge> const &edges, int N)
    {
        // resize the vector to N elements of type vector<int>
        adjList.resize(N);

        // add edges to the undirected graph
        for (auto &edge: edges)
        {
            adjList[edge.src].push_back(edge.dest);
            adjList[edge.dest].push_back(edge.src);
        }
    }
};

// Perform BFS recursively on graph
void recursiveBFS(Graph const &graph, queue<int> &q,
                    vector<bool> &discovered)
{
    if (q.empty())
        return;

    // pop front node from queue and print it
    int v = q.front();
    q.pop();
    cout << v << " ";

    // do for every edge (v -> u)
    for (int u : graph.adjList[v])
    {
        if (!discovered[u])
        {
            // mark it discovered and push it into queue
            discovered[u] = true;
            q.push(u);
        }
    }

    recursiveBFS(graph, q, discovered);
}

// Recursive C++ implementation of Breadth first search
int main()
{
    // vector of graph edges as per above diagram
    vector<Edge> edges = {
        {1, 2}, {1, 3}, {1, 4}, {2, 5}, {2, 6}, {5, 9},
        {5, 10}, {4, 7}, {4, 8}, {7, 11}, {7, 12}
        // vertex 0, 13 and 14 are single nodes
    };

    // Number of nodes in the graph
    int N = 15;

    // create a graph from edges
    Graph graph(edges, N);

    // stores vertex is discovered or not
    vector<bool> discovered(N, false);

    // create a queue used to do BFS
    queue<int> q;

    // Do BFS traversal from all undiscovered nodes to
    // cover all unconnected components of graph
    for (int i = 0; i < N; i++) {
        if (discovered[i] == false)
        {
            // mark source vertex as discovered
            discovered[i] = true;

            // push source vertex into the queue
            q.push(i);

            // start BFS traversal from vertex i
            recursiveBFS(graph, q, discovered);
        }
    }

    return 0;
}

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 

1 2 3 4 5 7 6 9 10 11 13

---

有人能解释我的怀疑是对的还是错的吗？如果有人能澄清我的疑问，这会很有帮助。

for循环从0到N（15），因此您目前有一种非常精细的方法来打印图形中的所有节点。例如，您可以使用union find来标识连接节点的群集。@Botje是图形的正确顺序（显示输出的方式）。我知道节点13，14来自N=15。它打印的顺序是否正确？在本例中看起来是这样，但通常不能依赖于邻接列表中节点的顺序，也不能依赖于图中节点的顺序。这就是为什么我说你最好用BFS做一些有用的事情，而不仅仅是打印节点。你正在执行4个BFS搜索，根为0、1、13和14。我还认为BFS的递归本质上是浪费——有一个队列，为什么需要（程序）堆栈？从我的观点来看，递归的主要优点是自由堆栈…@tucuxi如果我从代码中删除队列，它将如何成为BFS，然后有效地成为DFS，对吗？