不使用队列的广度优先搜索

广度优先搜索是一种图形遍历算法，可逐级遍历图形或树。在本文中，使用邻接表而不使用队列来实现图的BFS。

例子：

Input:
Output: BFS traversal = 2, 0, 3, 1
Explanation:
In the following graph, we start traversal from vertex 2. When we come to vertex 0, we look for all adjacent vertices of it. 2 is also an adjacent vertex of 0. If we don’t mark visited vertices, then 2 will be processed again and it will become a non-terminating process. Therefore, a Breadth-First Traversal of the following graph is 2, 0, 3, 1.

方法：可以使用给定来源的简单的广度优先遍历来解决此问题。该实现使用图的邻接表表示。

这里：

• STL向量容器用于存储BFS遍历所需的相邻节点列表和节点队列。
• DP阵列用于存储节点到源的距离。每次我们从一个节点移动到另一个节点时，距离都会增加1。如果到达节点的距离小于以前的距离，我们将更新DP [node]中存储的值。

下面是上述方法的实现：

// C++ implementation to demonstrate
// the above mentioned approach
  
#include 
using namespace std;
  
// Function to find the distance
// from the source to other nodes
void BFS(int curr, int N, vector& vis,
         vector& dp, vector& v,
         vector >& adj)
{
  
    while (curr <= N) {
  
        // Current node
        int node = v[curr - 1];
        cout << node << ", ";
  
        for (int i = 0; i < adj[node].size(); i++) {
  
            // Adjacent node
            int next = adj[node][i];
  
            if ((!vis[next])
                && (dp[next] < dp[node] + 1)) {
  
                // Stores the adjacent node
                v.push_back(next);
  
                // Increases the distance
                dp[next] = dp[node] + 1;
  
                // Mark it as visited
                vis[next] = true;
            }
        }
        curr += 1;
    }
}
  
// Function to print the distance
// from source to other nodes
void bfsTraversal(
    vector >& adj,
    int N, int source)
{
    // Initially mark all nodes as false
    vector vis(N + 1, false);
  
    // Initialize distance array with 0
    vector dp(N + 1, 0), v;
  
    v.push_back(source);
  
    // Initially mark the starting
    // source as 0 and visited as true
    dp = 0;
    vis = true;
  
    // Call the BFS function
    BFS(1, N, vis, dp, v, adj);
}
  
// Driver code
int main()
{
    // No. of nodes in graph
    int N = 4;
  
    // Creating adjacency list
    // for representing graph
    vector > adj(N + 1);
    adj[0].push_back(1);
    adj[0].push_back(2);
    adj[1].push_back(2);
    adj[2].push_back(0);
    adj[2].push_back(3);
    adj[3].push_back(3);
  
    // Following is BFS Traversal
    // starting from vertex 2
    bfsTraversal(adj, N, 2);
  
    return 0;
}

输出：

2, 0, 3, 1,

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