Graph Valid Tree
Given
n
nodes labeled from0
ton - 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
andedges = [[0, 1], [0, 2], [0, 3], [1, 4]]
, returntrue
.Given
n = 5
andedges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]
, returnfalse
.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 inedges
.
分析
这种验证有效的图都是一类题。无非就是几种方法,比如DFS。注意树有个特征就是一个child不能有两个parent, 根平时拓扑排序找环的的差别,当然由于这道题是无向图,不用考虑这方面。这题有几个要注意的地方,比如一个树只能有一个root,如果这个树存在多个root, 那肯定无效。
由于这道题是无向图,那么我们其实可以从任意一个点开始DFS,不用必须从root,因为无向图从任意一点都可以遍历到整块部分。添加edge的时候两个方向都添就可以了,另外就是还有个注意的地方就是如果在遍历neighbors的时候,自己parent不要遍历,否则直接返回false以为有环了。总结一句就是,这种无向图每个点遍历一次就可以了,否则就意味着有环。
如果这道题换成是有向图的话,解法见下文。
复杂度
time: O(V + E), space: O(h)
代码
public class Solution {
public boolean validTree(int n, int[][] edges) {
List<Integer>[] graph = new List[n];
for (int i = 0; i < n; i++) {
graph[i] = new ArrayList<Integer>();
}
for (int i = 0; i < edges.length; i++) {
graph[edges[i][0]].add(edges[i][1]);
graph[edges[i][1]].add(edges[i][0]);
}
boolean[] visited = new boolean[n];
if (!dfs(graph, 0, visited, -1)) {
return false;
}
// 检测是否存在多个root, 如是意味着图无效
for (int i = 0; i < n; i++) {
if (!visited[i]) {
return false;
}
}
return true;
}
private boolean dfs(List<Integer>[] graph, int i, boolean[] visited, int prev) {
// 发现环
if (visited[i]) {
return false;
}
visited[i] = true;
for (int num : graph[i]) {
if (prev != num && !dfs(graph, num, visited, i)) {
return false;
}
}
return true;
}
}
Graph Valid Tree(Directed Edges)
Given
n
nodes labeled from0
ton - 1
and a list of directed 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
andedges = [[0, 1], [0, 2], [0, 3], [1, 4]]
, returntrue
.Given
n = 5
andedges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]]
, returnfalse
.
Givenn = 5
andedges = [[0, 1], [1, 2], [3, 4]]
, returnfalse
.
分析
有向图要注意的地方不仅是不能有多个root存在,与上题不同的是需要从root开始遍历才能慢慢遍历到整个图,其他方面基本跟上题一样,遍历过程中一个点只能visit一次,否则即是那种一个child多个parent情况。
复杂度
time: O(V+E), space: O(h)
代码
public class Solution {
public boolean validTree(int n, int[][] edges) {
List<Integer>[] graph = new List[n];
Set<Integer> inDegree = new HashSet<>();
for (int i = 0; i < n; i++) {
graph[i] = new ArrayList<Integer>();
}
for (int i = 0; i < edges.length; i++) {
graph[edges[i][0]].add(edges[i][1]);
inDegree.add(edges[i][1]);
}
int root = 0;
int count = 0;
for (int i = 0; i < n; i++) {
if (!inDegree.contains(i)) {
root = i;
count++;
}
}
if (count != 1)
return false;
boolean[] visited = new boolean[n];
if (!dfs(graph, root, visited)) {
return false;
}
return true;
}
private boolean dfs(List<Integer>[] graph, int i, boolean[] visited) {
// 发现环或一child多个parents
if (visited[i]) {
return false;
}
visited[i] = true;
for (int num : graph[i]) {
if (!dfs(graph, num, visited)) {
return false;
}
}
return true;
}
}
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