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.

 Notice

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.

Example

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

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

Runtime: 213ms

class Solution {
public:
    /**
     * @param n an integer
     * @param edges a list of undirected edges
     * @return true if it's a valid tree, or false
     */
    bool validTree(int n, vector<vector<int>>& edges) {
        // Write your code here
        
        // use union find to solve this problem.
        // initialize the the union find array
        if (!n) return true;
        vector<int> uf;
        for (int i = 0; i < n; i++)
            uf.push_back(i);
        
        // change the parent of all pairs accordingly
        vector<bool> visited(n, false);
        for (int i = 0; i < edges.size(); i++) {
            if (visited[edges[i][0]] && visited[edges[i][1]] && uf[edges[i][0]] == uf[edges[i][1]]) return false;
22             updateFather(uf, edges[i][0], edges[i][1]); // The order of these lines are important!!!
            visited[edges[i][0]] = true;
            visited[edges[i][1]] = true;
        }
        
        // check whether have multiple fatheres
        int father = uf[0];
        for (int i = 1; i < n; i++) {
            if (uf[i] != father) return false;
        }
        return true;
    }
    
    void updateFather(vector<int>& uf, int from, int to) {
        int temp = uf[to];
        for (int i = 0; i < uf.size(); i++) {
            if (uf[i] == temp) uf[i] = uf[from];
        }
    }
};