LeetCode 261. Graph Valid Tree(判断图是否为树)
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原题网址:https://leetcode.com/problems/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
andedges = [[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.”
注意,这里的树是普通的树,不是二叉树!
思路:要判断一个图是否为树,首先要知道树的定义。一棵树必须具备如下特性:
(1)是一个全连通图(所有节点相通)
(2)无回路
其中(2)等价于:(3)图的边数=节点数-1
因此我们可以利用特性(1)(2)或者(1)(3)来判断。
方法一:广度优先搜索。要判断连通性,广度优先搜索法是一个天然的选择,时间复杂度O(n),空间复杂度O(n)。
public class Solution { public boolean validTree(int n, int[][] edges) { Map<Integer, Set<Integer>> graph = new HashMap<>(); for(int i=0; i<edges.length; i++) { for(int j=0; j<2; j++) { Set<Integer> pairs = graph.get(edges[i][j]); if (pairs == null) { pairs = new HashSet<>(); graph.put(edges[i][j], pairs); } pairs.add(edges[i][1-j]); } } Set<Integer> visited = new HashSet<>(); Set<Integer> current = new HashSet<>(); visited.add(0); current.add(0); while (!current.isEmpty()) { Set<Integer> next = new HashSet<>(); for(Integer node: current) { Set<Integer> pairs = graph.get(node); if (pairs == null) continue; for(Integer pair: pairs) { if (visited.contains(pair)) return false; next.add(pair); visited.add(pair); graph.get(pair).remove(node); } } current = next; } return visited.size() == n; }}
方法二:深度优先搜索,搜索目标是遍历全部节点。参考文章:http://buttercola.blogspot.com/2015/08/leetcode-graph-valid-tree.html
public class Solution { private boolean[] visited; private int visits = 0; private boolean isTree = true; private void check(int prev, int curr, List<Integer>[] graph) { if (!isTree) return; if (visited[curr]) { isTree = false; return; } visited[curr] = true; visits ++; for(int next: graph[curr]) { if (next == prev) continue; check(curr, next, graph); if (!isTree) return; } } public boolean validTree(int n, int[][] edges) { visited = new boolean[n]; List<Integer>[] graph = new List[n]; for(int i=0; i<n; i++) graph[i] = new ArrayList<>(); for(int[] edge: edges) { graph[edge[0]].add(edge[1]); graph[edge[1]].add(edge[0]); } check(-1, 0, graph); return isTree && visits == n; }}
方法三:按节点大小对边进行排序,原理类似并查集。
public class Solution { public boolean validTree(int n, int[][] edges) { if (edges.length != n-1) return false; Arrays.sort(edges, new Comparator<int[]>() { @Override public int compare(int[] e1, int[] e2) { return e1[0] - e2[0]; } }); int[] sets = new int[n]; for(int i=0; i<n; i++) sets[i] = i; for(int i=0; i<edges.length; i++) { if (sets[edges[i][0]] == sets[edges[i][1]]) return false; if (sets[edges[i][0]] == 0) { sets[edges[i][1]] = 0; } else if (sets[edges[i][1]] == 0) { sets[edges[i][0]] = 0; } else { sets[edges[i][1]] = sets[edges[i][0]]; } } return true; }}
方法四:Union-Find
public class Solution { public boolean validTree(int n, int[][] edges) { if (edges.length != n-1) return false; int[] roots = new int[n]; for(int i=0; i<n; i++) roots[i] = i; for(int i=0; i<edges.length; i++) { int root1 = root(roots, edges[i][0]); int root2 = root(roots, edges[i][1]); if (root1 == root2) return false; roots[root2] = root1; } return true; } private int root(int[] roots, int id) { if (id == roots[id]) return id; return root(roots, roots[id]); }}
参考文章:
http://blog.csdn.net/dm_vincent/article/details/7655764
http://www.elvisyu.com/graph-valid-tree-union-and-find/
http://blog.csdn.net/pointbreak1/article/details/48796691
http://buttercola.blogspot.com/2015/08/leetcode-graph-valid-tree.html
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