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:
Given n = 5 and edges = [[0, 1], [1, 2], [3, 4]], what should your return? Is this case a valid tree?
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.
Analysis:
First let’s try to understand the requirement of the valid tree. Tree is a graph which can’t have cycle, and moreover for a n nodes tree there are must have n - 1 edges. In this problem all the edges are undirected edges.
To solve the problem we need first to make the graph structure, because all the nodes are labeled from 0 to n - 1, so we could use vector
class Solution {
public:
bool validTree(int n, vector<pair<int, int>>& edges) {
vector<vector<int> > neighbors(n); // save graph
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))// need to record the parent because each edges will store in the two node's neighbors
return true;
return false;
}
};