310 Minimum Height Trees
For an 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 containsn
nodes which are labeled from0
ton - 1
. You will be given the numbern
and a list of undirectededges
(each edge is a pair of labels).
You can assume that no duplicate edges will appear inedges
. Since all edges are undirected,[0, 1]
is the same as[1, 0]
and thus will not appear together inedges
.
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]
Note:
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.”
The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.
一开始看到tag是bfs就死往那方面想,然后看了答案才发现自己对这个MHT的理解有误,一开始以为是最靠近叶子的那些点。写了半天还在1441的大数据TLE。其实这题跟topo sort有点像,也跟那题366 Find Leaves of Binary Tree 有点像。其实就是把叶子放到queue里,然后一层一层地剥。剥到剩下1个或2个叶子节点时就是答案了。这样的图里其实就有1或2棵那种MHT。
class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
if (edges == null || edges.length == 0 || edges[0].length == 0) {
List<Integer> res = new ArrayList<>();
for (int i = 0; i < n; i++) {
res.add(i);
}
return res;
}
// make the nodes
List<UndirectedGraph> graph = new ArrayList<>();
for (int i = 0; i < n; i++) {
graph.add(new UndirectedGraph(i));
}
// connect the edges
for (int i = 0; i < edges.length; i++) {
graph.get(edges[i][0]).neighbours.add(graph.get(edges[i][1]));
graph.get(edges[i][1]).neighbours.add(graph.get(edges[i][0]));
}
// put leaves in to queue first
Queue<UndirectedGraph> queue = new LinkedList<>();
for (UndirectedGraph node : graph) {
if (node.neighbours.size() == 1) {
queue.offer(node);
}
}
// remove leaves layer by layer until only 1 or 2 left
while (n > 2) {
int size = queue.size();
n = n - size;
for (int i = 0; i < size; i++) {
UndirectedGraph cur = queue.poll();
UndirectedGraph nei = cur.neighbours.get(0);
nei.neighbours.remove(cur);
// remmeber to put leaves into next round
if (nei.neighbours.size() == 1) {
queue.offer(nei);
}
}
}
List<Integer> res = new ArrayList<>();
for (UndirectedGraph node : queue) {
res.add(node.val);
}
return res;
}
}
class UndirectedGraph {
int val;
List<UndirectedGraph> neighbours = new ArrayList<>();
public UndirectedGraph(int value) {
val = value;
}
}
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