# Question

Formatted question description: https://leetcode.ca/all/261.html

You have a graph of n nodes labeled from 0 to n - 1. You are given an integer n and a list of edges where edges[i] = [ai, bi] indicates that there is an undirected edge between nodes ai and bi in the graph.

Return true if the edges of the given graph make up a valid tree, and false otherwise.

Example 1:

Input: n = 5, edges = [[0,1],[0,2],[0,3],[1,4]]
Output: true


Example 2:

Input: n = 5, edges = [[0,1],[1,2],[2,3],[1,3],[1,4]]
Output: false


Constraints:

• 1 <= n <= 2000
• 0 <= edges.length <= 5000
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• There are no self-loops or repeated edges.

# Algorithm

If it is a tree:

• all nodes must be connected, that is, it must be a connected graph
• and there must be no cycles

Use queue to assist the traversal. Here, instead of using a one-dimensional vector to mark whether the node has been visited, a HashSet is used. If a node is traversed and there is no HashSet in the HashSet, the HashSet is added. If it already exists, false is returned.

# Code

• import java.util.*;

public class Graph_Valid_Tree {

public class Solution {
/**
* @param n an integer
* @param edges a list of undirected edges
* @return true if it's a valid tree, or false
*/
public boolean validTree(int n, int[][] edges) {
if (n == 0) {
return false;
}

if (edges.length != n - 1) { // quick check
return false;
}

Map<Integer, Set<Integer>> graph = initializeGraph(n, edges);

// bfs
Set<Integer> isVisited = new HashSet<>();

queue.offer(0);
while (!queue.isEmpty()) {
int node = queue.poll();
for (Integer neighbor : graph.get(node)) {
if (isVisited.contains(neighbor)) { // cycle found
return false;
}
queue.offer(neighbor);
}
}

return (isVisited.size() == n); // if all nodes connected
}

private Map<Integer, Set<Integer>> initializeGraph(int n, int[][] edges) {

// node => its neibougher
Map<Integer, Set<Integer>> graph = new HashMap<>();
for (int i = 0; i < n; i++) {
graph.put(i, new HashSet<Integer>());
}

for (int i = 0; i < edges.length; i++) {
int u = edges[i][0];
int v = edges[i][1];
}

return graph;
}
}
}

############

class Solution {
private int[] p;

public boolean validTree(int n, int[][] edges) {
p = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
}
for (int[] e : edges) {
int a = e[0], b = e[1];
if (find(a) == find(b)) {
return false;
}
p[find(a)] = find(b);
--n;
}
return n == 1;
}

private int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
}

• // OJ: https://leetcode.com/problems/graph-valid-tree/
// Time: O(N + E)
// Space: O(N)
class UnionFind {
vector<int> id;
int cnt;
public:
UnionFind(int N) : id(N), cnt(N) {
iota(begin(id), end(id), 0);
}
int find(int x) {
return id[x] == x ? x : (id[x] = find(id[x]));
}
void connect(int a, int b) {
id[find(a)] = find(b);
--cnt;
}
bool connected(int a, int b) {
return find(a) == find(b);
}
int getCount() { return cnt; }
};
class Solution {
public:
bool validTree(int n, vector<vector<int>>& E) {
UnionFind uf(n);
for (auto &e : E) {
if (uf.connected(e[0], e[1])) return false;
uf.connect(e[0], e[1]);
}
return uf.getCount() == 1;
}
};

• class Solution:
def validTree(self, n: int, edges: List[List[int]]) -> bool:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]

p = list(range(n))
for a, b in edges:
if find(a) == find(b):
return False
p[find(a)] = find(b)
n -= 1
return n == 1

############

class Solution:
# @param {int} n an integer
# @param {int[][]} edges a list of undirected edges
# @return {boolean} true if it's a valid tree, or false
def validTree(self, n, edges):

def dfs(root, graph, visited, parent):
visited[root] = 1
for nbr in graph.get(root, []):
if nbr == parent:
continue
elif visited[nbr] != 0:
return False
if not dfs(nbr, graph, visited, root):
return False
visited[root] = 2
self.nodeVisited += 1
return True

visited = [0 for _ in range(n)]
graph = {}
self.nodeVisited = 0
for edge in edges:
start, end = edge[0], edge[1]
graph[start] = graph.get(start, []) + [end]
graph[end] = graph.get(end, []) + [start]

if dfs(0, graph, visited, -1) and self.nodeVisited == n:
return True
else:
return False


• func validTree(n int, edges [][]int) bool {
p := make([]int, n)
for i := range p {
p[i] = i
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
for _, e := range edges {
a, b := e[0], e[1]
if find(a) == find(b) {
return false
}
p[find(a)] = find(b)
n--
}
return n == 1
}

• /**
* @param {number} n
* @param {number[][]} edges
* @return {boolean}
*/
var validTree = function (n, edges) {
let p = new Array(n);
for (let i = 0; i < n; ++i) {
p[i] = i;
}
function find(x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
for (const [a, b] of edges) {
if (find(a) == find(b)) {
return false;
}
p[find(a)] = find(b);
--n;
}
return n == 1;
};