For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.
Format The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).
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.
Example 1 :
Input: n = 4, edges = [[1, 0], [1, 2], [1, 3]]
0
|
1
/ \
2 3
Output: [1]
Example 2 :
Input: n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]
0 1 2
\ | /
3
|
4
|
5
Output: [3, 4]
- Method 1: bfs, TLE
class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer> result = new ArrayList<>();
if(n <= 0 || edges == null) return result;
if(n == 1){
result.add(0);
return result;
}
Map<Integer, List<Integer>> map = new HashMap<>();
for(int i = 0; i < n; i++)
map.put(i, new ArrayList<Integer>());
for(int[] arr: edges){
map.get(arr[0]).add(arr[1]);
map.get(arr[1]).add(arr[0]);
}
int min = Integer.MAX_VALUE;
LABEL:
for(int i = 0; i < n; i++){
boolean[] visited = new boolean[n];
List<Integer> list = map.get(i);
LinkedList<Integer> queue = new LinkedList<>();
queue.offer(i);
int count = 0;
while(!queue.isEmpty()){
count++;
if(count > min) continue LABEL;
int size = queue.size();
for(int j = 0; j < size; j++){
int cur = queue.poll();
if(visited[cur]) continue;
visited[cur] = true;
queue.addAll(map.get(cur));
}
}
if(count == min)
result.add(i);
else if(count < min){
min = count;
result.clear();
result.add(i);
}
}
return result;
}
}- Method 2: 类似拓扑
- 先找到所有入度为1的，这些节点为叶子结点。
- 不断从邻接表中删除对应的节点，直到最后剩下的结点，即为根节点。
class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer> leaves = new ArrayList<>();
if(n <= 0 || edges == null) return leaves;
if(n == 1){
leaves.add(0);
return leaves;
}
List<Integer>[] map = new List[n];
for(int i = 0; i < n; i++)
map[i] = new ArrayList<Integer>();
for(int[] arr: edges){
map[arr[0]].add(arr[1]);
map[arr[1]].add(arr[0]);
}
for(int i = 0; i < n; i++)
if(map[i].size() == 1){
leaves.add(i);
}
while(n > 2){
n -= leaves.size();
List<Integer> newLeaves = new ArrayList<>();
for(int i : leaves){
Integer cur = map[i].get(0);
map[cur].remove(map[cur].indexOf(i));
if(map[cur].size() == 1)
newLeaves.add(cur);
}
leaves = newLeaves;
}
return leaves;
}
}- BFS, AC, almost TLE
class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer>[] bag = new List[n];
for(int i = 0; i < n; i++){
bag[i] = new LinkedList<Integer>();
}
for(int[] pair : edges){
bag[pair[0]].add(pair[1]);
bag[pair[1]].add(pair[0]);
}
int min = Integer.MAX_VALUE;
List<Integer> result = new LinkedList<>();
for(int i = 0; i < n; i++){
boolean[] visited = new boolean[n];
LinkedList<Integer> queue = new LinkedList<>();
queue.add(i);
int count = 0;
while(!queue.isEmpty()){
int size = queue.size();
for(int j = 0; j < size; j++){
Integer v = queue.poll();
visited[v] = true;
for(Integer node : bag[v]){
if(!visited[node]){
queue.add(node);
}
}
}
count++;
}
if(count == min) result.add(i);
if(count < min){
result.clear();
result.add(i);
min = count;
}
}
return result;
}
}- Calculate the indegree of each node, if indegree is 1, add to queue.
- once the remained node number is small or equal to 2, break the queue.
class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer> result = new LinkedList<>();
if(n <= 2){
for(int i = 0; i < n; i++) result.add(i);
return result;
}
int[] count = new int[n];
boolean[] visited = new boolean[n];
List<Integer>[] bag = new List[n];
for(int i = 0; i < n; i++) bag[i] = new LinkedList<>();
for(int[] pair : edges){
count[pair[0]]++;
bag[pair[0]].add(pair[1]);
count[pair[1]]++;
bag[pair[1]].add(pair[0]);
}
LinkedList<Integer> queue = new LinkedList<>();
int v = n;
for(int i = 0; i < n; i++){
if(count[i] == 1){
queue.add(i);
count[i]--;
}
}
while(!queue.isEmpty() && v > 2){
int size = queue.size();
v -= size;
for(int i = 0; i < size; i++){
int val = queue.poll();
visited[val] = true;
List<Integer> list = bag[val];
for(int node: list){
count[node]--;
if(count[node] == 1) queue.add(node);
}
}
}
for(int i = 0; i < n; i++){
if(!visited[i]) result.add(i);
}
return result;
}
}