leetcode-java/src/GraphValidTree261.java at master · fluency03/leetcode-java · GitHub

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/**
 * 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.
 *
 * Example 1:
 * Input: n = 5, and edges = [[0,1], [0,2], [0,3], [1,4]]
 * Output: true
 *
 * Example 2:
 * Input: n = 5, and edges = [[0,1], [1,2], [2,3], [1,3], [1,4]]
 * Output: false
 * 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.
 */

publicclassGraphValidTree261 {
// 1. construct Graph
// 2. detection cycle -- DFS
// 3. check not more than one tree
publicbooleanvalidTree(intn, int[][] edges) {
if (n == 0) returnfalse;
if (n == 1) returntrue;
Map<Integer, Set<Integer>> graph = constructGraph(n, edges);
boolean[] visited = newboolean[n];

// if cycle detected, return false
if (!isValid(graph, visited, 0, -1)) returnfalse;
for (inti=0; i<n; i++) {
if (!visited[i]) returnfalse;
 }
returntrue;
 }

// return false if cycle detected
privatebooleanisValid(Map<Integer, Set<Integer>> graph, boolean[] visited, intcurr, intparent) {
if (visited[curr]) {
returnfalse;
 }

visited[curr] = true;
if (!graph.containsKey(curr)) returntrue;
for (inti: graph.get(curr)) {
if (i != parent && !isValid(graph, visited, i, curr)) returnfalse;
 }
returntrue;
 }

privateMap<Integer, Set<Integer>> constructGraph(intn, int[][] edges) {
Map<Integer, Set<Integer>> graph = newHashMap<>();
for (inti=0; i<n; i++) {
graph.put(i, newHashSet<>());
 }
for (int[] edge: edges) {
graph.get(edge[0]).add(edge[1]);
graph.get(edge[1]).add(edge[0]);
 }
returngraph;
 }


publicbooleanvalidTree2(intn, int[][] edges) {
DisjointSetdjs = newDisjointSet(n);
for (int[] edge: edges) {
intx = djs.find(edge[0]);
inty = djs.find(edge[1]);
if (x == y) returnfalse;
djs.union(x, y);
 }
introot = djs.find(0);
for (inti=1; i<n; i++) {
if (djs.find(i) != root) returnfalse;
 }
returntrue;
 }

classDisjointSet {
int[] parent;
int[] rank;

publicDisjointSet(intn) {
this.parent = newint[n];
for (inti=0; i<n; i++) this.parent[i] = i;
this.rank = newint[n];
 }

publicintfind(intx) {
if (parent[x] != x) {
parent[x] = find(parent[x]);
 }
returnparent[x];
 }

publicvoidunion(intx, inty) {
intxx = find(x);
intyy = find(y);
if (rank[xx] > rank[yy]) {
parent[yy] = xx;
 } elseif (rank[xx] < rank[yy]) {
parent[xx] = yy;
 } else {
parent[xx] = yy;
rank[yy]++;
 }
 }
 }


/**
 * https://leetcode.com/problems/graph-valid-tree/discuss/69018/AC-Java-Union-Find-solution
 */
publicbooleanvalidTree3(intn, int[][] edges) {
// initialize n isolated islands
int[] nums = newint[n];
Arrays.fill(nums, -1);

// perform union find
for (inti = 0; i < edges.length; i++) {
intx = find(nums, edges[i][0]);
inty = find(nums, edges[i][1]);

// if two vertices happen to be in the same set
// then there's a cycle
if (x == y) returnfalse;

// union
nums[x] = y;
 }

returnedges.length == n - 1;
 }

intfind(intnums[], inti) {
if (nums[i] == -1) returni;
returnfind(nums, nums[i]);
 }

}