[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:

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

.

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);
}
};