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Graph Valid Tree in C++
Suppose we have n nodes they are labeled from 0 to n-1 and a list of undirected edges [u,v], We have to define a function to check whether these edges make up a valid tree or not.
So, if the input is like n = 5, and edges = [[0,1], [0,2], [0,3], [1,4]], then the output will be true
To solve this, we will follow these steps −
Define a function dfs(), this will take node, par, graph, and another array called visited,
if visited[node] is same as 1, then −
return true
if visited[node] is same as 2, then −
return false
visited[node] := 2
ret := true
for initialize i := 0, when i < size of graph[node], update (increase i by 1), do −
if graph[node, i] is not equal to par, then −
ret := ret AND dfs(graph[node, i], node, graph, visited)
visited[node] := 1
return ret
From the main method do the following −
Define an array visited of size n and fill this with 0.
Define a list of lists called graph of size n
for initialize i := 0, when i < size of edges, update (increase i by 1), do −
u := edges[i, 0], v := edges[i, 1]
insert v at the end of graph[u]
insert u at the end of graph[v]
if dfs(0, -1, graph, visited) is false, then −
return false
for initialize i := 0, when i < n, update (increase i by 1), do −
if visited[i] is zero, then −
return false
return true
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool dfs(int node, int par, vector <int< graph[], vector <int<& visited){
if (visited[node] == 1)
return true;
if (visited[node] == 2)
return false;
visited[node] = 2;
bool ret = true;
for (int i = 0; i < graph[node].size(); i++) {
if (graph[node][i] != par)
ret &= dfs(graph[node][i], node, graph, visited);
}
visited[node] = 1;
return ret;
}
bool validTree(int n, vector<vector<int<>& edges) {
vector<int< visited(n, 0);
vector<int< graph[n];
for (int i = 0; i < edges.size(); i++) {
int u = edges[i][0];
int v = edges[i][1];
graph[u].push_back(v);
graph[v].push_back(u);
}
if (!dfs(0, -1, graph, visited))
return false;
for (int i = 0; i < n; i++) {
if (!visited[i])
return false;
}
return true;
}
};
main(){
Solution ob;
vector<vector<int<> v = {{0,1},{0,2},{0,3},{1,4}};
cout << (ob.validTree(5,v));
}Input
5, {{0,1},{0,2},{0,3},{1,4}}Output
1
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