Related Articles

Traverse graph in lexicographical order of nodes using DFS

• Last Updated : 20 Jul, 2021

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 using namespace std; // Function to traverse the graph in// lexicographical order using DFSvoid LexiDFS(map >& G,             char S, map& 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 nodesvoid CreateGraph(int N, int M, int S,                 char Edges[][2]){    // Store all the adjacent nodes    // of each node of a graph    map > 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 vis;     // Function Call    LexiDFS(G, S, vis);} // Driver Codeint 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 approachimport java.util.*; class Graph{ // Function to traverse the graph in// lexicographical order using DFSstatic void LexiDFS(HashMap> G,            char S, HashMap 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 nodesstatic void CreateGraph(int N, int M, char S,                        char[][] Edges){         // Store all the adjacent nodes    // of each node of a graph    HashMap> G = new HashMap<>();     // Traverse Edges[][2] array    for(int i = 0; i < M; i++)    {        if (G.containsKey(Edges[i][0]))        {            Set temp = G.get(Edges[i][0]);            temp.add(Edges[i][1]);            G.put(Edges[i][0], temp);        }        else        {            Set temp = new HashSet<>();            temp.add(Edges[i][1]);            G.put(Edges[i][0], temp);        }    }         // Check if a node is already visited or not    HashMap vis = new HashMap<>();     LexiDFS(G, S, vis);} // Driver codepublic 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 approachG = [[] for i in range(300)]vis = [0 for i in range(300)] # Function to traverse the graph in# lexicographical order using DFSdef 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 nodesdef 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 Codeif __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


Output:
a p b c y l m z g h i

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up