261 Graph Valid Tree

1. Question

Givennnodes labeled from0ton - 1and 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:
Givenn = 5andedges = [[0, 1], [0, 2], [0, 3], [1, 4]], returntrue.
Givenn = 5andedges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], returnfalse.
Note: you can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges.

2. Implementation

(1) BFS
class Solution {
public boolean validTree(int n, int[][] edges) {
List<Set<Integer>> adjList = new ArrayList<>();
for (int i = 0; i < n; i++) {
adjList.add(new HashSet<>());
}
for (int[] edge : edges) {
adjList.get(edge[0]).add(edge[1]);
adjList.get(edge[1]).add(edge[0]);
}
boolean[] visited = new boolean[n];
Queue<Integer> queue = new LinkedList<>();
queue.add(0);
while (!queue.isEmpty()) {
int curNode = queue.remove();
// found loop
if (visited[curNode]) {
return false;
}
visited[curNode] = true;
for (int nextNode : adjList.get(curNode)) {
queue.add(nextNode);
adjList.get(nextNode).remove(curNode);
}
}
for (boolean e : visited) {
if (!e) {
return false;
}
}
return true;
}
}
(2) DFS
class Solution {
public boolean validTree(int n, int[][] edges) {
List<Set<Integer>> adjList = new ArrayList<>();
for (int i = 0; i < n; i++) {
adjList.add(new HashSet<>());
}
for (int[] edge : edges) {
adjList.get(edge[0]).add(edge[1]);
adjList.get(edge[1]).add(edge[0]);
}
boolean[] visited = new boolean[n];
// Check if graph has cycle
if (hasCycle(0, visited, adjList, -1)) {
return false;
}
// Check if graph is connected
for (int i = 0; i < n; i++) {
if (!visited[i]) {
return false;
}
}
return true;
}
public boolean hasCycle(int node, boolean[] visited, List<Set<Integer>> adjList, int parent) {
visited[node] = true;
for (int nextNode : adjList.get(node)) {
// (1) If nextNode is visited but it is not the parent of the curNode, then there is cycle
// (2) If nextNode is not visited but we still find the cycle later on, return true;
if ((visited[nextNode] && parent != nextNode) || (!visited[nextNode] && hasCycle(nextNode, visited, adjList, node))) {
return true;
}
}
return false;
}
}
(3) Union Find
class Solution {
public boolean validTree(int n, int[][] edges) {
UnionFind uf = new UnionFind(n);
for (int[] edge : edges) {
// Find Loop
if (!uf.union(edge[0], edge[1])) {
return false;
}
}
// Make sure the graph is connected
return uf.count == 1;
}
class UnionFind {
int[] sets;
int[] size;
int count;
public UnionFind(int n) {
sets = new int[n];
size = new int[n];
count = n;
for (int i = 0; i < n; i++) {
sets[i] = i;
size[i] = 1;
}
}
public int find(int node) {
while (node != sets[node]) {
node = sets[node];
}
return node;
}
public boolean union(int i, int j) {
int node1 = find(i);
int node2 = find(j);
if (node1 == node2) {
return false;
}
if (size[node1] < size[node2]) {
sets[node1] = node2;
size[node2] += size[node1];
}
else {
sets[node2] = node1;
size[node1] += size[node2];
}
--count;
return true;
}
}
}

3. Time & Space Complexity

BFS: 时间复杂度O(n), 空间复杂度O(n)
DFS: 时间复杂度O(n), 空间复杂度O(n)
Union Find: 时间复杂度O(n), 空间复杂度O(n)