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:
Given n = 4
, edges = [[1, 0], [1, 2], [1, 3]]
0 | 1 / \ 2 3
return [1]
Example 2:
Given n = 6
, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]
0 1 2 \ | / 3 | 4 | 5
return [3, 4]
Note:
(1) 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.”
(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.
Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.
提示说一棵树中最多有几颗MHT,答案是2,这个是可以证明的,但是博主并不会- -依照这个思路我们可以每次删去度数为1的叶结点,直到树中只剩1-2个结点即为所求,代码参考的https://discuss.leetcode.com/topic/48519/java-bfs-like-ac-solution-using-user-defined-graph-node-class,我觉得比我自己写的要清晰:
public List<Integer> findMinHeightTrees(int n, int[][] edges) { Map<Integer, Node> graph = new HashMap<>(); for (int i = 0; i < n; i++) { graph.put(i, new Node(i)); } for (int[] edge : edges) { graph.get(edge[0]).neighbors.add(graph.get(edge[1])); graph.get(edge[1]).neighbors.add(graph.get(edge[0])); graph.get(edge[0]).degree++; graph.get(edge[1]).degree++; } Queue<Node> queue = new LinkedList<>(); for (int index : graph.keySet()) { if (graph.get(index).degree == 1) { queue.offer(graph.get(index)); } } while (n > 2) { int size = queue.size(); for (int i = 0; i < size; i++) { Node leaf = queue.poll(); Node neighbor = leaf.neighbors.iterator().next(); neighbor.neighbors.remove(leaf);//remove leaf from its neighbor's adj list graph.remove(leaf.label);//remove leaf self n--; if (--neighbor.degree == 1) { queue.offer(neighbor); } } } return new ArrayList<>(graph.keySet()); } } class Node { int label; int degree; Set<Node> neighbors; public Node(int index) { label = index; neighbors = new HashSet<>(); }