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
For a ‘proper’ tree (see below), it’s the same thing, at least by most definitions. Like Wikipedia, for example: Breadth-first See also: Breadth-first search Trees can also be traversed in level-order, … … a breadth-first (level-order) traversal … Well, at least level-order traversal is the same as breadth-first traversal. There are many reasons to traverse … Read more
As a caveat, remember that there can be exponentially many shortest paths between two nodes in a graph. Any algorithm for this will potentially take exponential time. That said, there are a few relatively straightforward algorithms that can find all the paths. Here’s two. BFS + Reverse DFS When running a breadth-first search over a … Read more
Just build one level at a time, e.g.: class Node(object): def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right def traverse(rootnode): thislevel = [rootnode] while thislevel: nextlevel = list() for n in thislevel: print n.value, if n.left: nextlevel.append(n.left) if n.right: nextlevel.append(n.right) print thislevel = nextlevel t = Node(1, Node(2, Node(4, … Read more
Breadth first search Queue<TreeNode> queue = new LinkedList<BinaryTree.TreeNode>() ; public void breadth(TreeNode root) { if (root == null) return; queue.clear(); queue.add(root); while(!queue.isEmpty()){ TreeNode node = queue.remove(); System.out.print(node.element + ” “); if(node.left != null) queue.add(node.left); if(node.right != null) queue.add(node.right); } }
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
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.
Consider a graph like this: A—(3)—–B | | \-(1)-C–(1)/ The shortest path from A to B is via C (with a total weight of 2). A normal BFS will take the path directly from A to B, marking B as seen, and A to C, marking C as seen. At the next stage, propagating from … Read more