Breadth first search python

I've been working on building different data types and applying various sorting algorithms to them. I'm currently working on a breadth-first search on a binary search tree. My code is pretty much the same as what you'll find everywhere online, yet it consistently prints my values twice, and I'm now mind boggled. Any guidance would very much appreciated. # Remove (dequeue) function for Queue class def remove(self, current=''): if current == '': current = self.tail # start looping/recurring to find the front of the queue if current.next: if current.next.next: current = current.next return self.remove(current) # side note - I'm doubting the usefulness of recursion here else: n = current.next.value current.next = None self.size -= 1 return n elif current == self.tail: if current.value: n = current.value current = None self.size -= 1 # print("Queue is now empty - returning final number") return n else: return "Queue is already empty." else: raise ValueError("mind boggling error...") #never raised # Add (enqueue) function for my queue: def add(self,value): n = Node(value) # create the new node we're adding n.next = self.tail # point the new node to link to the old tail self.tail = n # now have the tail point to the new node self.size += 1 # Breadth First Search function for the Binary Search Tree def bfs(self): """ This is currently VERY inefficient from a memory standpoint, as it adds a subtree for EACH node...right? or is it just the names/addresses of the objects being stored...

Breadth-first search (BFS or Level Order Traversal) is a method of traversing a tree or graph data structure. BFS uses the The BFS algorithm starts at the root node and travels through every child node at the current level before moving to the next level. BFS is an extremely common pattern used to solve algorithmic puzzles in technical interviews and competitive programming sites such as leetcode and hackerrank. Space and time complexity Time complexity: O(n) where n is the number of nodes. Space complexity: O(b) where b is maximum breadth. Use case examples: BFS is often used for finding the shortest path or solving for a series of choices which can result in a winning or losing state. Used in: Dijkstra's Algorithm, Ford-Fulkerson Algorithm Animated solution def bfs(self, root=None): if root is None: return queue = [root] while len(queue) > 0: cur_node = queue.pop(0) if cur_node.left is not None: queue.append(cur_node.left) if cur_node.right is not None: queue.append(cur_node.right) # end # initialize tree and start bfs traversal root = Node(4) root.insert_nodes([2, 1, 3, 6, 5, 7], root) root.bfs(root=root) class Node: def __init__(self, val: int): self.left = None self.right = None self.val = val def __repr__(self): return str(self.val) def insert_node(self, val): if self.val is not None: if val self.val: if self.right is None: self.right = Node(val) else: self.right.insert_node(val) @staticmethod def insert_nodes(vals: list, root): for i in vals: root.insert_node(i) de...

I have a tree as input to the breadth first search and I want to know as the algorithm progresses at which level it is? # Breadth First Search Implementation graph = for item in graph.keys(): mark[item] = 0 queue, output = [],[] # Initialize an empty queue with the source node and mark it as explored queue.append(source) mark[source] = 1 output.append(source) # while queue is not empty while queue: # remove the first element of the queue and call it vertex vertex = queue[0] queue.pop(0) # for each edge from the vertex do the following for vrtx in graph[vertex]: # If the vertex is unexplored if mark[vrtx] == 0: queue.append(vrtx) # mark it as explored mark[vrtx] = 1 # and append it to the queue output.append(vrtx) # fill the output vector return output print breadth_first_search(graph, 'A') It takes tree as an input graph, what I want is, that at each iteration it should print out the current level which is being processed. Actually, we don't need an extra queue to store the info on the current depth, nor do we need to add null to tell whether it's the end of current level. We just need to know how many nodes the current level contains, then we can deal with all the nodes in the same level, and increase the level by 1 after we are done processing all the nodes on the current level. int level = 0; Queue queue = new LinkedList(); queue.add(root); while(!queue.isEmpty()) You don't need to use extra queue or do any complicated calculation to achieve what you want to do. This i...

How does Level Order Traversal work? The main idea of level order traversal is to traverse all the nodes of a lower level before moving to any of the nodes of a higher level. This can be done in any of the following ways: • the naive one (finding the height of the tree and traversing each level and printing the nodes of that level) • efficiently using a queue. Level Order Traversal (Naive approach): Find height of tree. Then for each level, run a recursive function by maintaining current height. Whenever the level of a node matches, print that node. Output Level Order traversal of binary tree is 1 2 3 4 5 Time Complexity: O(N 2), where N is the number of nodes in the skewed tree. Auxiliary Space: O(1) If the recursion stack is considered the space used is O(N). Level Order Traversal using We need to visit the nodes in a lower level before any node in a higher level, this idea is quite similar to that of a queue. Push the nodes of a lower level in the queue. When any node is visited, pop that node from the queue and push the child of that node in the queue. This ensures that the node of a lower level are visited prior to any node of a higher level. Below is the Implementation of the above approach:

and a • Start (0, 0), and • Initialize a boolean (0, 0) as visited. • Declare a function isValid() to check if the cell coordinates are valid or not, i.e lies within the boundaries of the given Matrix and are unvisited or not. • Iterate while the • • Move to its adjacent cells that are not visited. • Mark them visited and enqueue them into the queue. Note: Direction vectors are used to traverse the adjacent cells of a given cell in a given order. For example (x, y) is a cell whose adjacent cells (x – 1, y), (x, y + 1), (x + 1, y), (x, y – 1) need to be traversed, then it can be done using the direction vectors (-1, 0), (0, 1), (1, 0), (0, -1) in the up, left, down and right order. Below is the implementation of the above approach:

by Sachin Malhotra Finding Shortest Paths using Breadth First Search Do you know the amount of global air traffic in 2017? Do you know what the rise has been for air traffic over the past several years ? Well, lets look at some statistics. Source: According to the That’s a lot of passengers and a lot of flights occupying the air space on a daily basis across the world. Since there are hundreds and thousands of these flights all around the globe, there are bound to be different routes with multiple stops in between from one place to another. Every flight has a source and destination of its own and a standard economy seat price associated with it. Let’s leave out the fancy business class tickets and extra leg room and what not! In such a scenario, it is too confusing to choose what flight would be the best one if we want to go from one place to another. Let’s see the number of flight options Every flight has a Details hyperlink with it, so we searched for that and found 119 total flights. 119 total flights are being offered. Then there appears a pop up on the website saying that there are other websites that might be offering similar flights at even cheaper rates. ? So many websites and uncountable flights for just a single source and destination. As a developer, if I want to solve this problem, I would build a system to efficiently address the following queries: • Total number of destinations reachable (with a max number of stops) from my current location, and also list tho...

Breadth First Search (BFS) has been discussed in Adjacency matrix representation: In adjacency matrix representation of a graph, the matrix mat[][] of size n*n (where n is the number of vertices) will represent the edges of the graph where mat[i][j] = 1 represents that there is an edge between the vertices i and j while mat[i][j] = 0 represents that there is no edge between the vertices i and j. first, before moving on to the solution. Approach: • Create a matrix of size n*n where every element is 0 representing there is no edge in the graph. • Now, for every edge of the graph between the vertices i and j set mat[i][j] = 1. • After the adjacency matrix has been created and filled, find the BFS traversal of the graph as described in Below is the implementation of the above approach: