[LeetCode] 310. Minimum Height Trees 最小高度树

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]

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.

Java:

public List<Integer> findMinHeightTrees(int n, int[][] edges) {
    List<Integer> result = new ArrayList<Integer>();
    if(n==0){
        return result;
    }
    if(n==1){
        result.add(0);
        return result;
    }
 
    ArrayList<HashSet<Integer>> graph = new ArrayList<HashSet<Integer>>();
    for(int i=0; i<n; i++){
        graph.add(new HashSet<Integer>());
    }
 
    for(int[] edge: edges){
        graph.get(edge[0]).add(edge[1]);
        graph.get(edge[1]).add(edge[0]);
    }
 
    LinkedList<Integer> leaves = new LinkedList<Integer>();
    for(int i=0; i<n; i++){
        if(graph.get(i).size()==1){
            leaves.offer(i);
        }
    }
 
    if(leaves.size()==0){
        return result;
    }
 
    while(n>2){
        n = n-leaves.size();
 
        LinkedList<Integer> newLeaves = new LinkedList<Integer>();
 
        for(int l: leaves){
            int neighbor = graph.get(l).iterator().next();
            graph.get(neighbor).remove(l);
            if(graph.get(neighbor).size()==1){
                newLeaves.add(neighbor);
            }
        }
 
        leaves = newLeaves;
    }
 
    return leaves;
}　　

Python:

class Solution(object):
    def findMinHeightTrees(self, n, edges):
        """
        :type n: int
        :type edges: List[List[int]]
        :rtype: List[int]
        """
        if n == 1:
            return [0]
 
        neighbors = collections.defaultdict(set)
        for u, v in edges:
            neighbors[u].add(v)
            neighbors[v].add(u)
 
        pre_level, unvisited = [], set()
        for i in xrange(n):
            if len(neighbors[i]) == 1:  # A leaf.
                pre_level.append(i)
            unvisited.add(i)
 
        # A graph can have 2 MHTs at most.
        # BFS from the leaves until the number
        # of the unvisited nodes is less than 3.
        while len(unvisited) > 2:
            cur_level = []
            for u in pre_level:
                unvisited.remove(u)
                for v in neighbors[u]:
                    if v in unvisited:
                        neighbors[v].remove(u)
                        if len(neighbors[v]) == 1:
                            cur_level.append(v)
            pre_level = cur_level
 
        return list(unvisited)

C++:

// Time:  O(n)
// Space: O(n)
 
class Solution {
public:
    vector<int> findMinHeightTrees(int n, vector<pair<int, int>>& edges) {
        if (n == 1) {
            return {0};
        }
 
        unordered_map<int, unordered_set<int>> neighbors;
        for (const auto& e : edges) {
            int u, v;
            tie(u, v) = e;
            neighbors[u].emplace(v);
            neighbors[v].emplace(u);
        }
 
        vector<int> pre_level, cur_level;
        unordered_set<int> unvisited;
        for (int i = 0; i < n; ++i) {
            if (neighbors[i].size() == 1) {  // A leaf.
                pre_level.emplace_back(i);
            }
            unvisited.emplace(i);
        }
 
        // A graph can have 2 MHTs at most.
        // BFS from the leaves until the number
        // of the unvisited nodes is less than 3.
        while (unvisited.size() > 2) {
            cur_level.clear();
            for (const auto& u : pre_level) {
                unvisited.erase(u);
                for (const auto& v : neighbors[u]) {
                    if (unvisited.count(v)) {
                        neighbors[v].erase(u);
                        if (neighbors[v].size() == 1) {
                            cur_level.emplace_back(v);
                        }
                    }
                }
            }
            swap(pre_level, cur_level);
        }
 
        vector<int> res(unvisited.begin(), unvisited.end());
        return res;
    }
};

　　

　　

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