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.
Notice
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.
Have you met this question in a real interview? Yes 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.
這個題目的重點的部分是 判斷graph進行遍歷時，進出的個數與原有的點的個數的對比
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) { // Write your code here if (n == 0) { return false; } if (n - 1 != edges.length) { return false; } Map<Integer, Set<Integer>> map = initializeGraph(n, edges); Queue<Integer> queue = new LinkedList<>(); Set<Integer> set = new HashSet<>(); queue.offer(0); set.add(0); int visit = 0; while (!queue.isEmpty()) { int node = queue.poll(); visit++; for (int root : map.get(node)) { if (set.contains(root)) { continue; } queue.offer(root); set.add(root); } } return visit == n; } PRivate Map<Integer, Set<Integer>> initializeGraph(int n, int[][] edges) { 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]; graph.get(u).add(v); graph.get(v).add(u); } return graph; }}新聞熱點
疑難解答
