【LeetCode】310. Minimum Height Trees
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 :
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]
題解：可知最後解只能是一個或兩個點，如果三個點以上，以其中一個點為根，其他兩點必定在其子樹上，因為演算法是一層一層的去掉外面的葉節點，所以最後剩下三個點時外面兩個點肯定要去掉，因為一層一層的剝開保證了對每個根節點每次都減掉一層也就是最後的根節點來說高度不變這樣最後只能有一個或兩個
class Solution {
public:
vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
vector<int> ans;
if (n == 1) return {0};
vector<unordered_set<int>> tree(n);
for(int i=0;i<edges.size();i++){
tree[edges[i].first].insert(edges[i].second);
tree[edges[i].second].insert(edges[i].first);
}
queue<int> q;
for(int i=0;i<n;i++){
if(tree[i].size()==1) q.push(i);
}
while(n>2){
int size=q.size();
n-=size;
for(int i=0;i<size;i++){
int x=q.front();
q.pop();
for(auto t:tree[x]){
tree[t].erase(x);
if(tree[t].size()==1) q.push(t);
}
}
}
while(!q.empty()){
ans.push_back(q.front());
q.pop();
}
return ans;
}
};