LeetCode 261. Graph Valid Tree(判斷圖是否為樹)
阿新 • • 發佈:2019-01-25
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]]
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,這裡的樹是普通的樹,不是二叉樹!
思路:要判斷一個圖是否為樹,首先要知道樹的定義。一棵樹必須具備如下特性:
(1)是一個全連通圖(所有節點相通)
(2)無迴路
其中(2)等價於:(3)圖的邊數=節點數-1
因此我們可以利用特性(1)(2)或者(1)(3)來判斷。
方法一:廣度優先搜尋。要判斷連通性,廣度優先搜尋法是一個天然的選擇,時間複雜度O(n),空間複雜度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; } }
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:按節點大小對邊進行排序,原理類似並查集。
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]);
}
}
參考文章: