[LeetCode] Graph Valid Tree

Problem Description:

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. 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 inedges.

---

As suggested by the hint, just check for cycle and connectedness in the graph. Both of these can be done via DFS.

The code is as follows.

class Solution {
public:
    bool validTree(int n, vector<pair<int, int>>& edges) {
        vector<vector<int>> neighbors(n);
        for (auto e : edges) {
            neighbors[e.first].push_back(e.second);
            neighbors[e.second].push_back(e.first);
        }
        vector<bool> visited(n, false);
        if (hasCycle(neighbors, 0, -1, visited))
            return false;
        for (bool v : visited)
            if (!v) return false; 
        return true;
    }
private:
    bool hasCycle(vector<vector<int>>& neighbors, int kid, int parent, vector<bool>& visited) {
        if (visited[kid]) return true;
        visited[kid] = true;
        for (auto neigh : neighbors[kid])
            if (neigh != parent && hasCycle(neighbors, neigh, kid, visited))
                return true;
        return false;
    }
};