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 exactlyone 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.
class Solution {
public:
vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
vector<int> res;
if(n==1)
{
res.push_back(0);
return res;
}
unordered_map<int ,unordered_set<int> > adj;
for(auto &entry : edges)
{
adj[entry.first].insert(entry.second);
adj[entry.second].insert(entry.first);
}
vector<int> leaves;
for(auto &entry : adj)
{
if(adj[entry.first].size()==1)
{
leaves.push_back(entry.first);
}
}
while(n>2)
{
n -= leaves.size();
vector<int> nextLeaves;
for(int idx=0; idx<leaves.size(); idx++)
{
int nextNode = *(adj[leaves[idx]].begin());
adj[leaves[idx]].erase(nextNode);
adj[nextNode].erase(leaves[idx]);
if(adj[nextNode].size()==1)
{
nextLeaves.push_back(nextNode);
}
}
leaves = nextLeaves;
}
return leaves;
}
};