[LeetCode] Graph Valid Tree

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?
2. According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.”

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.

首先要判断是不是只有一个连通分支，如果不是肯定不是树，如果只有一个连通分支，再判断分支内有没有环，如果没有环，那么就是树。求连通分支个数用并查集，判断是否有环用DFS。

class Solution {
public:
    int findFather(vector<int> &father, int x) {
        while (x != father[x]) x = father[x];
        return x;
    }
    bool dfs(vector<vector<int>> &graph, vector<bool> &visit, int u, int f) {
        visit[u] = true;
        for (auto v : graph[u]) if (v != f) {
            if (visit[v]) return false;
            else if (!dfs(graph, visit, v, u)) return false;
        }
        return true;
    }
    bool validTree(int n, vector<pair<int, int>>& edges) {
        vector<int> father(n);
        for (int i = 0; i < n; ++i) father[i] = i;
        for (auto edge : edges) {
            int fa = findFather(father, edge.first), fb = findFather(father, edge.second);
            if (fa < fb) father[fb] = fa;
            else father[fa] = fb;
        };
        int cnt = 0;
        for (int i = 0; i < n; ++i) {
            if (i == father[i]) ++cnt;
            if (cnt > 1) return false;
        }
        vector<vector<int>> graph(n);
        vector<bool> visit(n, false);
        for (auto edge : edges) {
            int u = edge.first, v = edge.second;
            graph[u].push_back(v);
            graph[v].push_back(u);
        }
        return dfs(graph, visit, 0, -1);
    }
};