LeetCode 261. Graph Valid Tree（判断图是否为树）
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 = 5andedges = [[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> graph = new HashMap<>();
for(int i=0; i 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 visited = new HashSet<>();
Set current = new HashSet<>();
visited.add(0);
current.add(0);
while (!current.isEmpty()) {
Set next = new HashSet<>();
for(Integer node: current) {
Set 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[] 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[] graph = new List[n];
for(int i=0; i();
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() {
@Override
public int compare(int[] e1, int[] e2) {
return e1[0] - e2[0];
}
});
int[] sets = new int[n];
for(int i=0; i 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