breadth-first-search

I mean on a specific level, NOT up to that specific level. Could someone please check my modified BFS algorithm? (most of which is taken from Wikipedia) Queue levelorder(root, levelRequested){ int currentLevel = 0; q = empty queue q.enqueue(root) while not q.empty do{ if(currentLevel==levelRequested) return q; node := q.dequeue() visit(node) if(node.left!=null || node.right!=null){ currentLevel++; if node.left ≠ null q.enqueue(node.left) if node.right ≠ null q.enqueue(node.right) } } } I think a recursive solution would be much more concise: /* * node - node being visited * clevel - current

'Length' of a path is the number of edges in the path. Given a source and a destination vertex, I want to find the number of paths form the source vertex to the destination vertex of given length k. We can visit each vertex as many times as we want, so if a path from a to b goes like this: a -> c -> b -> c -> b it is considered valid. This means there can be cycles and we can go through the destination more than once. Two vertices can be connected by more than one edge. So if vertex a an vertex b are connected by two edges, then the paths , a -> b via edge 1 and a -> b via edge 2 are

The following question was found in Sedgewick and Wayne book about algorithms in java: 4.2.19 Topological sort and BFS. Explain why the following algorithm does not necessarily produce a topological order: Run BFS, and label the vertices by increasing distance to their respective source. I was trying to prove it finding a counter example. But, everytime I try, I get a topological order. I mean, I don't understand why this not work: If the source of a vertex comes before it, why don't we have a topological order? I think to prove it, we need to find a vertex that its source comes before, but I

I am currently trying to traverse all paths from source to destination in a graph which uses adjacency matrix. I have been trying to do it in BFS way.Thanks for the help. I am getting only one path. How do I get to print other paths as well ? public class AllPossiblePaths { static int v; static ArrayList<Integer> adj[]; public AllPossiblePaths(int v) { this.v = v; adj = new ArrayList[v]; for (int i = 0; i < v; i++) { adj[i] = new ArrayList<>(); } } // add edge from u to v public static void addEdge(int u, int v) { adj[u].add(v); } public static void findpaths(int source, int destination) {

void traverse(Node* root) { queue<Node*> q; Node* temp_node= root; while(temp_node) { cout<<temp_node->value<<endl; if(temp_node->left) q.push(temp_node->left); if(temp_node->right) q.push(temp_node->right); if(!q.empty()) { temp_node = q.front(); q.pop(); } else temp_node = NULL; } } The above posted code is my level order traversal code. This code works fine for me but One thing I dont like is I am explicitly initializing temp_node = NULL or I use break. But it does not seem to be a good code to me. Is there a neat implementation than this or how can I make this code better? void traverse

This is an interview question I think of a solution. It uses queue. public Void BFS() { Queue q = new Queue(); q.Enqueue(root); Console.WriteLine(root.Value); while (q.count > 0) { Node n = q.DeQueue(); if (n.left !=null) { Console.Writeln(n.left); q.EnQueue(n.left); } if (n.right !=null) { Console.Writeln(n.right); q.EnQueue(n.right); } } } Can anything think of better solution than this, which doesn't use Queue? CodeFusionMobile Level by level traversal is known as Breadth-first traversal. Using a Queue is the proper way to do this. If you wanted to do a depth first traversal you would use a

I was revising single source shortest path algorithms and in the video, the teacher mentions that BFS/DFS can't be used directly for finding shortest paths in a weighted graph (I guess everyone knows this already) and said to work out the reason on your own. I was wondering the exact reason/explanation as to why it can't be used for weighted graphs. Is it due to the weights of the edges or anything else ? Can someone explain me as I feel a little confused. PS: I went through this question and this question. Consider a graph like this: A---(3)-----B | | \-(1)-C--(1)/ The shortest path from A to