261. Graph Valid Tree

问题描述：

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.

Example 1:

Input: n = 5, and edges = [[0,1], [0,2], [0,3], [1,4]]
Output: true

Example 2:

Input: n = 5, and edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]
Output: false

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.

 

解题思路：

若一个无向图为一棵树，则图中无环

我们可以用Union Find来确认图中无环

注意：

　　1. 本题目中给出节点个数，若边的个数为空，说明图中只有节点，若节点个数为0或1，则为一棵有效的树，否则不能构成树

　　2. 在对每条边进行Union操作后，需要对每一个点进行find来确定每一个点连接到同一个根上。

　　　　一开始取根为  int root = find(parents, 0);  而非 int root = parents[0]

 

代码：

class Solution {
public:
    bool validTree(int n, vector<pair<int, int>>& edges) {
        if(edges.empty())
            return n < 2;
        vector<int> parents(n);
        for(int i = 0; i < n; i++)
            parents[i] = i;
        for(auto e : edges){
            if(!Union(parents, e.first, e.second))
                return false;
        }
        int root = find(parents, 0);
        for(int i = 1; i < n; i++){
            if(find(parents, i) != root)
                return false;
        }
        return true;
    }
private:
    int find(vector<int> &parents, int a){
        if(parents[a] == a)
            return a;
        return parents[a] = find(parents, parents[a]);
    }
    bool Union(vector<int> &parents, int a, int b){
        int rootA = find(parents, a);
        int rootB = find(parents, b);
        if(rootA != rootB){
            parents[rootB] = rootA;
            return true;
        }
        return false;
    }
};