LeetCode 261. Graph Valid Tree

Given n nodes labeled from 0 to n - 1 and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.

For example:

Given n = 5 and edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return true.

Given n = 5 and edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return false.

Hint:

1. Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], what should your return? Is this case a valid tree? Show More Hint 

Note: you can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.

Method 1: Union-Find Algorithm

#include <vector>#include <iostream>using namespace std;// this is actually to use union-find.bool find(vector<int>& id, int p, int q) {  return id[p] == id[q];}int root(vector<int>& id, int p) {  while(p != id[p]) {    p = id[p];  }  return p;}void unite(vector<int>& id, int p, int q) {  int root_p = root(id, p);  int root_q = root(id, q);  id[root_p] = root_q;}bool isValidTree(int n, vector< vector<int> >& edges) {  if(n < 0 || (n - 1) != edges.size()) return false;  vector<int> id(n);  for(int i = 0; i < n; ++i) {    id[i] = i;  }  for(int i = 0; i < edges.size(); ++i) {    if(find(id, edges[i][0], edges[i][1])) {      return false;    } else {      unite(id, edges[i][0], edges[i][1]);    }  }  return true;}int main(void) {  vector< vector<int> > edges{{0, 1}, {0, 2}, {0, 3}, {1, 4}};  vector< vector<int> > edges_1{{0, 1}, {1, 2}, {2, 3}, {1, 3}, {1, 4}};  cout << isValidTree(5, edges) << endl;  cout << isValidTree(5, edges_1) << endl;}

If could't think up Union-Find method, it is also Ok to to use DFS to detect cycle. If cycle found, that is not a valid graph tree.

Step:

 1: first build up hashmap hashmap[first].push_back(second), hashmap[second].push_back(first)

 2: Make all the node unvisited.

 3: Loop all the node, if found an unvisited node, using dfs to detect cycle. Note: here we need to pass the parent node in to dfs function

    bool dfs(nodesMap, currNode, parentNode, visited) {

         visited[curreNode] = true;

        for(v : edges) {

             if(unvisited) {if(dfs(nodesMap, v, currentNode, visited)) return true;}

             else if( *v != parentNode) return ture // cycle detected

        }

      return false // there is no cycle.

    }

   if(dfs) return false // cycle in the undirected graph, that is not a valid tree.

Beside checking whether there is a cycle, we also need to check whether the given nodes forms a forest.