Minimum Height Trees
题目详情:
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]
解题方法:
代码详情:
public:
vector<int> findMinHeightTrees(int n, vector<pair<int, int> >& edges) {
vector<int> result;
if (n == 0) {
return result;
} else if (n == 1) {
result.push_back(0);
return result;
}
vector<unordered_set<int> > graph(n);
// make graph
for (int i = 0; i < edges.size(); i++) {
graph[edges[i].first].insert(edges[i].second);
graph[edges[i].second].insert(edges[i].first);
}
// find leaf node, push in queue
queue<int> q;
for (int i = 0; i < n; i++) {
if (graph[i].size() == 1) {
q.push(i);
}
}
while (n > 2) {
int size = q.size();
n -= size;
for (int i = 0; i < size; i++) {
int temp = q.front();
q.pop();
for (auto node:graph[temp]) {
graph[node].erase(temp);
if (graph[node].size() == 1) {
q.push(node);
}
}
}
}
while (!q.empty()) {
result.push_back(q.front());
q.pop();
}
return result;
}
};