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Check if a given graph is tree or not
In this problem, one undirected graph is given, we have to check the graph is tree or not. We can simply find it by checking the criteria of a tree. A tree will not contain a cycle, so if there is any cycle in the graph, it is not a tree.
We can check it using another approach, if the graph is connected and it has V-1 edges, it could be a tree. Here V is the number of vertices in the graph.
Input and Output
Input: The adjacency matrix. 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 1 1 1 1 0 Output: The Graph is a tree
Algorithm
isCycle(u, visited, parent)
Input: The start vertex u, the visited list to mark visited or not, the parent vertex.
Output: True if there is a cycle in the graph.
Begin mark u as visited for all vertex v which are adjacent with u, do if v is visited, then if isCycle(v, visited, u) = true, then return true else if v ≠ parent, then return true done return false End
isTree(graph)
Input: The undirected graph.
Output: True when the graph is a tree.
Begin define a visited array to mark which node is visited or not initially mark all node as unvisited if isCycle(0, visited, φ) is true, then //the parent of starting vertex is null return false if the graph is not connected, then return false return true otherwise End
Example
#include<iostream>
#define NODE 5
using namespace std;
int graph[NODE][NODE] = {
{0, 1, 1, 1, 0},
{1, 0, 1, 0, 0},
{1, 1, 0, 0, 0},
{1, 0, 0, 0, 1},
{0, 0, 0, 1, 0}
};
bool isCycle(int u, bool visited[], int parent) {
visited[u] = true; //mark v as visited
for(int v = 0; v<NODE; v++) {
if(graph[u][v]) {
if(!visited[v]) { //when the adjacent node v is not visited
if(isCycle(v, visited, u)) {
return true;
}
} else if(v != parent) { //when adjacent vertex is visited but not parent
return true; //there is a cycle
}
}
}
return false;
}
bool isTree() {
bool *vis = new bool[NODE];
for(int i = 0; i<NODE; i++)
vis[i] = false; //initialize as no node is visited
if(isCycle(0, vis, -1)) //check if there is a cycle or not
return false;
for(int i = 0; i<NODE; i++) {
if(!vis[i]) //if there is a node, not visited by traversal, graph is not connected
return false;
}
return true;
}
int main() {
if(isTree())
cout << "The Graph is a Tree.";
else
cout << "The Graph is not a Tree.";
}Output
The Graph is a Tree.
Updated on: 16-Jun-2020
3K+ Views
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