The time complexities for BFS and DFS are just O(|E|), or in your case, O(m). In a binary tree, m is equal to n-1 so the time complexity is equivalent to O(|V|). m refers to the total number of edges, not the average number of adjacent edges per vertex.
DFS does not necessarily yield shortest paths in an undirected graph. BFS would be the correct choice here. As an example, consider a graph formed by taking the corners of a triangle and connecting them. If you try to find the shortest path from one node to another using DFS, then you will get the … Read more
Finding articulation vertices is an application of DFS. In a nutshell, Apply DFS on a graph. Get the DFS tree. A node which is visited earlier is a “parent” of those nodes which are reached by it and visited later. If any child of a node does not have a path to any of the … Read more
A DFS without recursion is basically the same as BFS – but use a stack instead of a queue as the data structure. The thread Iterative DFS vs Recursive DFS and different elements order handles with both approaches and the difference between them (and there is! you will not traverse the nodes in the same … Read more
The short answer is yes, there are situations in which A* is not the best algorithm to solve a problem. However, there are a number of ways to assess what constitutes the best algorithm for finding a solution. If you are considering best in terms of performance of multiple searches from a single source to … Read more
Pre-order is one type of DFS. There are three types of depth-first traversal: pre-order, in-order, and post-order. Check out here for more info.
E is the set of all edges in the graph, as G={V,E}. So, |E| is count of all edges in the graph. This alone should be enough to convince you that the overall complexity can’t be |V| times |E|, since we are not iterating over all the edges in the graph for each vertex. In … Read more
Queue can be generally thought as horizontal in structure i.e, breadth/width can be attributed to it – BFS, whereas Stack is visualized as a vertical structure and hence has depth – DFS.
In a depth-first search, you begin at some node in the graph and continuously explore deeper and deeper into the graph while you can find new nodes that you haven’t yet reached (or until you find the solution). Any time the DFS runs out of moves, it backtracks to the latest point where it could … Read more
Answering my question: The graph has a cycle if and only if there exists a back edge. A back edge is an edge that is from a node to itself (selfloop) or one of its ancestor in the tree produced by DFS forming a cycle. Both approaches above actually mean the same. However, this method … Read more