Answers for "dfs gfg"

9

dfs python

###############
#The Algorithm (In English):

# 1) Pick any node. 
# 2) If it is unvisited, mark it as visited and recur on all its 
#    adjacent nodes. 
# 3) Repeat until all the nodes are visited, or the node to be 
#    searched is found.


# The graph below (declared as a Python dictionary)
# is from the linked website and is used for the sake of
# testing the algorithm. Obviously, you will have your own
# graph to iterate through.
graph = {
    'A' : ['B','C'],
    'B' : ['D', 'E'],
    'C' : ['F'],
    'D' : [],
    'E' : ['F'],
    'F' : []
}

visited = set() # Set to keep track of visited nodes.


##################
# The Algorithm (In Code)

def dfs(visited, graph, node):
    if node not in visited:
        print (node)
        visited.add(node)
        for neighbour in graph[node]:
            dfs(visited, graph, neighbour)
            
# Driver Code to test in python yourself.
# Note that when calling this, you need to
# call the starting node. In this case it is 'A'.
dfs(visited, graph, 'A')

# NOTE: There are a few ways to do DFS, depending on what your
# variables are and/or what you want returned. This specific
# example is the most fleshed-out, yet still understandable,
# explanation I could find.
Posted by: Guest on October-05-2020
2

DFS in c++

#include <bits/stdc++.h>
using namespace std;
 

class Graph {
    int V; 
 
 
    list<int>* adj;
 
  
    void DFSUtil(int v, bool visited[]);
 
public:
    Graph(int V);
 
    void addEdge(int v, int w);
 
  
    void DFS(int v);
};
 
Graph::Graph(int V)
{
    this->V = V;
    adj = new list<int>[V];
}
 
void Graph::addEdge(int v, int w)
{
    adj[v].push_back(w); 
}
 
void Graph::DFSUtil(int v, bool visited[])
{
   
    visited[v] = true;
    cout << v << " ";
 
   
    list<int>::iterator i;
    for (i = adj[v].begin(); i != adj[v].end(); ++i)
        if (!visited[*i])
            DFSUtil(*i, visited);
}
 

void Graph::DFS(int v)
{
   
    bool* visited = new bool[V];
    for (int i = 0; i < V; i++)
        visited[i] = false;
 
 
    DFSUtil(v, visited);
}
 

int main()
{
  
    Graph g(4);
    g.addEdge(0, 1);
    g.addEdge(0, 2);
    g.addEdge(1, 2);
    g.addEdge(2, 0);
    g.addEdge(2, 3);
    g.addEdge(3, 3);
 
    cout << "Following is Depth First Traversal"
            " (starting from vertex 2) \n";
    g.DFS(2);
 
    return 0;
}
Posted by: Guest on December-08-2020
1

iterative dfs in tree

import java.util.Stack; 

class Node 
{ 
    int data; 
    Node left, right; 
  
    public Node(int item) 
    { 
        data = item; 
        left = right = null; 
    } 
} 

@SuppressWarnings("unchecked")
class BinaryTree
{
    Node root;
    public static void inOrderTraversal(Node root)
    {
      Stack<Node> stack = new Stack();
      Node curr = root;
      while (!stack.empty() || curr != null)
      {
        if (curr != null)
        {
          stack.push(curr);
          curr = curr.left;
        }
        else
        {
          curr = stack.pop();
          System.out.print(curr.data + " ");
          curr = curr.right;
        }
      }
    }

}
Posted by: Guest on September-27-2020

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