Traverse graph in lexicographical order of nodes using DFS

Last Updated : 15 Mar, 2023

Given a graph, G consisting of N nodes, a source S, and an array Edges[][2] of type {u, v} that denotes that there is an undirected edge between node u and v, the task is to traverse the graph in lexicographical order using DFS.

Examples:

Input: N = 10, M = 10, S = ‘a’, Edges[][2] = { { ‘a’, ‘y’ }, { ‘a’, ‘z’ }, { ‘a’, ‘p’ }, { ‘p’, ‘c’ }, { ‘p’, ‘b’ }, { ‘y’, ‘m’ }, { ‘y’, ‘l’ }, { ‘z’, ‘h’ }, { ‘z’, ‘g’ }, { ‘z’, ‘i’ } }
Output: a p b c y l m z g h i

Explanation:
For the first level visit the node and print it:

Similarly visited the second level node p which is lexicographical smallest as:

Similarly visited the third level for node p in lexicographical order as:

Now the final traversal is shown in the below image and labelled as increasing order of number:

Input: N = 6, S = ‘a’, Edges[][2] = { { ‘a’, ‘e’ }, { ‘a’, ‘d’ }, { ‘e’, ‘b’ }, { ‘e’, ‘c’ }, { ‘d’, ‘f’ }, { ‘d’, ‘g’ } }
Output: a d f g e b c

Approach: Follow the steps below to solve the problem:

Initialize a map, say G to store all the adjacent nodes of a node according to lexicographical order of the nodes.
Initialize a map, say vis to check if a node is already traversed or not.
Traverse the Edges[][2] array and store all the adjacent nodes of each node of the graph in G.
Finally, traverse the graph using DFS and print the visited nodes of the graph.

Below is the implementation of the above approach:

C++

// C++ program  for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to traverse the graph in
// lexicographical order using DFS
void LexiDFS(map<char, set<char> >& G,
             char S, map<char, bool>& vis)
{
    // Mark S as visited nodes
    vis[S] = true;
 
    // Print value of visited nodes
    cout << S << " ";
 
    // Traverse all adjacent nodes of S
    for (auto i = G[S].begin();
         i != G[S].end(); i++) {
 
        // If i is not visited
        if (!vis[*i]) {
 
            // Traverse all the nodes
            // which is connected to i
            LexiDFS(G, *i, vis);
        }
    }
}
 
// Utility Function to traverse graph
// in lexicographical order of nodes
void CreateGraph(int N, int M, int S,
                 char Edges[][2])
{
    // Store all the adjacent nodes
    // of each node of a graph
    map<char, set<char> > G;
 
    // Traverse Edges[][2] array
    for (int i = 0; i < M; i++) {
 
        // Add the edges
        G[Edges[i][0]].insert(
            Edges[i][1]);
    }
 
    // Check if a node is already
    // visited or not
    map<char, bool> vis;
 
    // Function Call
    LexiDFS(G, S, vis);
}
 
// Driver Code
int main()
{
    int N = 10, M = 10, S = 'a';
    char Edges[M][2]
        = { { 'a', 'y' }, { 'a', 'z' },
            { 'a', 'p' }, { 'p', 'c' }, 
            { 'p', 'b' }, { 'y', 'm' }, 
            { 'y', 'l' }, { 'z', 'h' }, 
            { 'z', 'g' }, { 'z', 'i' } };
 
    // Function Call
    CreateGraph(N, M, S, Edges);
 
    return 0;
}

Java

// Java program for above approach
import java.util.*;
 
class Graph{
 
// Function to traverse the graph in
// lexicographical order using DFS
static void LexiDFS(HashMap<Character, Set<Character>> G,
            char S, HashMap<Character, Boolean> vis)
{
     
    // Mark S as visited nodes
    vis.put(S, true);
 
    // Print value of visited nodes
    System.out.print(S + " ");
 
    // Traverse all adjacent nodes of S
    if (G.containsKey(S))
    {
        for(char i : G.get(S)) 
        {
             
            // If i is not visited
            if (!vis.containsKey(i) || !vis.get(i)) 
            {
                 
                // Traverse all the nodes
                // which is connected to i
                LexiDFS(G, i, vis);
            }
        }
    }
}
 
// Utility Function to traverse graph
// in lexicographical order of nodes
static void CreateGraph(int N, int M, char S,
                        char[][] Edges)
{
     
    // Store all the adjacent nodes
    // of each node of a graph
    HashMap<Character, Set<Character>> G = new HashMap<>();
 
    // Traverse Edges[][2] array
    for(int i = 0; i < M; i++)
    {
        if (G.containsKey(Edges[i][0]))
        {
            Set<Character> temp = G.get(Edges[i][0]);
            temp.add(Edges[i][1]);
            G.put(Edges[i][0], temp);
        }
        else
        {
            Set<Character> temp = new HashSet<>();
            temp.add(Edges[i][1]);
            G.put(Edges[i][0], temp);
        }
    }
     
    // Check if a node is already visited or not
    HashMap<Character, Boolean> vis = new HashMap<>();
 
    LexiDFS(G, S, vis);
}
 
// Driver code
public static void main(String[] args) 
{
    int N = 10, M = 10;
    char S = 'a';
 
    char[][] Edges = { { 'a', 'y' }, { 'a', 'z' },
                       { 'a', 'p' }, { 'p', 'c' },
                       { 'p', 'b' }, { 'y', 'm' },
                       { 'y', 'l' }, { 'z', 'h' },
                       { 'z', 'g' }, { 'z', 'i' } };
 
    // Function Call
    CreateGraph(N, M, S, Edges);
}
}
 
// This code is contributed by hritikrommie

Python3

# Python3 program  for the above approach
G = [[] for i in range(300)]
vis = [0 for i in range(300)]
 
# Function to traverse the graph in
# lexicographical order using DFS
def LexiDFS(S):
    global G, vis
     
    # Mark S as visited nodes
    vis[ord(S)] = 1
 
    # Prvalue of visited nodes
    print (S,end=" ")
 
    # Traverse all adjacent nodes of S
    for i in G[ord(S)]:
        # If i is not visited
        if (not vis[i]):
            # Traverse all the nodes
            # which is connected to i
            LexiDFS(chr(i))
 
# Utility Function to traverse graph
# in lexicographical order of nodes
def CreateGraph(N, M, S, Edges):
    global G
    # Store all the adjacent nodes
    # of each node of a graph
 
    # Traverse Edges[][2] array
    for i in Edges:
        # Add the edges
        G[ord(i[0])].append(ord(i[1]))
        G[ord(i[0])] = sorted(G[ord(i[0])])
 
    # Function Call
    LexiDFS(S)
 
# Driver Code
if __name__ == '__main__':
    N = 10
    M = 10
    S = 'a'
    Edges=[ ['a', 'y' ],[ 'a', 'z' ],
           [ 'a', 'p' ],[ 'p', 'c' ],
           [ 'p', 'b' ],[ 'y', 'm' ],
           [ 'y', 'l' ],[ 'z', 'h' ],
           [ 'z', 'g' ],[ 'z', 'i' ] ]
 
    # Function Call
    CreateGraph(N, M, S, Edges);
 
# This code is contributed by mohitkumar29.

Javascript

<script>
 
// JavaScript program  for the above approach
 
let G = new Array(300).fill(0).map(() => [])
let vis = new Array(300).fill(0)
 
// Function to traverse the graph in
// lexicographical order using DFS
function LexiDFS(S) {
    // Mark S as visited nodes
    vis[S.charCodeAt(0)] = 1
 
    // Prvalue of visited nodes
    document.write(S + " ")
 
    // Traverse all adjacent nodes of S
    for (let i of G[S.charCodeAt(0)]) {
        // If i is not visited
        if (!vis[i]) {
            // Traverse all the nodes
            // which is connected to i
            LexiDFS(String.fromCharCode(i))
        }
    }
}
 
// Utility Function to traverse graph
// in lexicographical order of nodes
function CreateGraph(N, M, S, Edges) {
    // Store all the adjacent nodes
    // of each node of a graph
 
    // Traverse Edges[][2] array
    for (let i of Edges) {
        // Add the edges
        G[i[0].charCodeAt(0)].push(i[1].charCodeAt(0))
         
        G[i[0].charCodeAt(0)] = 
        G[i[0].charCodeAt(0)].sort((a, b) => a - b)
    }
 
    // Function Call
    LexiDFS(S)
}
 
// Driver Code
let N = 10
let M = 10
let S = 'a'
let Edges = [['a', 'y'], ['a', 'z'],
['a', 'p'], ['p', 'c'],
['p', 'b'], ['y', 'm'],
['y', 'l'], ['z', 'h'],
['z', 'g'], ['z', 'i']]
 
// Function Call
CreateGraph(N, M, S, Edges);
 
// This code is contributed by _saurabh_jaiswal
 
</script>

C#

// C# program  for the above approach
using System;
using System.Collections.Generic;
 
class Graph {
    // Function to traverse the graph in
    // lexicographical order using DFS
    static void LexiDFS(List<int>[] G, int S, bool[] vis) {
        // Mark S as visited nodes
        vis[S] = true;
 
        // Print value of visited nodes
        Console.Write(Convert.ToChar(S) + " ");
 
        // Traverse all adjacent nodes of S
        foreach (int i in G[S]) {
            // If i is not visited
            if (!vis[i]) {
                // Traverse all the nodes
                // which is connected to i
                LexiDFS(G, i, vis);
            }
        }
    }
 
    // Utility Function to traverse graph
    // in lexicographical order of nodes
    static void CreateGraph(int N, int M, char S, char[][] Edges) {
        // Store all the adjacent nodes
        // of each node of a graph
        List<int>[] G = new List<int>[300];
        for (int i = 0; i < G.Length; i++) {
            G[i] = new List<int>();
        }
 
        // Traverse Edges[][2] array
        for (int i = 0; i < M; i++) {
            // Add the edges
            G[Edges[i][0]].Add(Edges[i][1]);
 
            G[Edges[i][0]].Sort();
        }
 
        // Check if a node is already visited or not
        bool[] vis = new bool[300];
 
        LexiDFS(G, S, vis);
    }
 
    // Driver code
    static void Main(string[] args) {
        int N = 10;
        int M = 10;
        char S = 'a';
 
        char[][] Edges = { 
            new char[] { 'a', 'y' }, 
            new char[] { 'a', 'z' },
            new char[] { 'a', 'p' }, 
            new char[] { 'p', 'c' },
            new char[] { 'p', 'b' }, 
            new char[] { 'y', 'm' },
            new char[] { 'y', 'l' }, 
            new char[] { 'z', 'h' },
            new char[] { 'z', 'g' }, 
            new char[] { 'z', 'i' }
        };
 
        // Function Call
        CreateGraph(N, M, S, Edges);
    }
}
 
// This code is contributed by Jay

Output:

a p b c y l m z g h i

Time Complexity: O(N * log(N))
Auxiliary Space: O(N)

Suggest improvement

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Traverse graph in lexicographical order of nodes using DFS

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