思路:按层剪掉 leaf nodes,当最终只剩一个或者两个 nodes,即为题解
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
LinkedList<Integer> res = new LinkedList<>();
if (n == 1) {
res.add(0);
return res;
}
int[] indegrees = new int[n];
HashMap<Integer, List<Integer>> map = new HashMap<>();
for (int[] edge: edges) {
indegrees[edge[0]]++;
indegrees[edge[1]]++;
if (!map.containsKey(edge[0])) {
map.put(edge[0], new LinkedList<Integer>());
}
map.get(edge[0]).add(edge[1]);
if (!map.containsKey(edge[1])) {
map.put(edge[1], new LinkedList<Integer>());
}
map.get(edge[1]).add(edge[0]);
}
Queue<Integer> queue = new LinkedList<>();
for (int i = 0; i < n; i++) {
if (indegrees[i] == 1) {
queue.offer(i);
}
}
int visitedSize = 0;
while (!queue.isEmpty()) {
if (visitedSize >= n - 2) {
break;
}
int size = queue.size();
for (int i = 0; i < size; i++) {
int node = queue.poll();
indegrees[node]--;
visitedSize++;
for (Integer neighbor: map.get(node)) {
if (indegrees[neighbor] == 0) {
continue;
}
indegrees[neighbor]--;
if (indegrees[neighbor] == 1) {
queue.offer(neighbor);
}
}
}
}
while (!queue.isEmpty()) {
res.add(queue.poll());
}
return res;
}
可以使用 HashMap<Integer, Set<Integer>> map
,这样就不用使用 indegrees array 了。思路是一致的,Graph Topological sort based on BFS.
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