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;
}
};

  

  

类似题目:

Course Schedule II

Course Schedule

Clone Graph

All LeetCode Questions List 题目汇总

[LeetCode] 310. Minimum Height Trees 最小高度树的更多相关文章

  1. [LeetCode] 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 gra ...

  2. [LeetCode] 310. Minimum Height Trees 解题思路

    For a undirected graph with tree characteristics, we can choose any node as the root. The result gra ...

  3. leetcode@ [310] Minimum Height Trees

    For a undirected graph with tree characteristics, we can choose any node as the root. The result gra ...

  4. 310. Minimum Height Trees -- 找出无向图中以哪些节点为根,树的深度最小

    For a undirected graph with tree characteristics, we can choose any node as the root. The result gra ...

  5. 【LeetCode】310. Minimum Height Trees 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 BFS 相似题目 参考资料 日期 题目地址:http ...

  6. 310. Minimum Height Trees

    For a undirected graph with tree characteristics, we can choose any node as the root. The result gra ...

  7. [LeetCode] 310. Minimum Height Trees_Medium tag: BFS

    For a undirected graph with tree characteristics, we can choose any node as the root. The result gra ...

  8. [Swift]LeetCode310. 最小高度树 | Minimum Height Trees

    For an undirected graph with tree characteristics, we can choose any node as the root. The result gr ...

  9. 最小高度的树 Minimum Height Trees

    2018-09-24 12:01:38 问题描述: 问题求解: 毫无疑问的一条非常好的题目,采用的解法是逆向的BFS,也就是从叶子节点开始遍历,逐步向中心靠拢,最终留下的叶子节点就是答案. publi ...

随机推荐

  1. ARTS-week3

    Algorithm 给定一个排序数组,你需要在原地删除重复出现的元素,使得每个元素只出现一次,返回移除后数组的新长度.不要使用额外的数组空间,你必须在原地修改输入数组并在使用 O(1) 额外空间的条件 ...

  2. 深度学习Keras框架笔记之TimeDistributedDense类

    深度学习Keras框架笔记之TimeDistributedDense类使用方法笔记 例: keras.layers.core.TimeDistributedDense(output_dim,init= ...

  3. MSDS596 Homework 9

    MSDS596 Homework 9 (Due in class on Nov 14) Fall 2017Notes. The lowest grade among all twelve homewo ...

  4. linux学习8 运维基本功-Linux获取命令使用帮助详解

    一.Linux基础知识 1.人机交互界面: a.GUI b.CLI:[login@hostname workdir]# COMMAND 2.命令知识 通用格式:# COMMAND  OPTIONS A ...

  5. tox python项目虚拟环境管理自动化测试&&构建工具

    tox 是一个方便的工具,可以帮助我们管理python 的虚拟环境,同时可以进行项目自动测试以及构建 tox 如何工作的 说明 从上图我们也可以看出如何在我们项目中使用tox 参考资料 https:/ ...

  6. IIS 站点配置文件

    IIS 站点配置文件  C:/Windows/System32/inetsrv/config/applicationHost.config 配置文件示例: <system.application ...

  7. 【后缀数组】【LuoguP4248】 [AHOI2013]差异

    题目链接 题目描述 给定一个长度为 n 的字符串 S,令 Ti 表示它从第 i 个字符开始的后缀.求 \(\sum_{1\le i <j\le n}len(T_i)+len(T_j)-2*lcp ...

  8. js无限轮播算法中干掉if判断

    无限轮播在网页应用中经常见到,这其中算法各有千秋,在学习算法分析一书中发现自增取余方法可以干掉一些不必要的if判断,具体代码如下: var arr= [1,2,3,4,5,6,7,8]; var in ...

  9. 微信小程序电影模板

    [外链图片转存失败(img-STw401rR-1565101469846)(https://upload-images.jianshu.io/upload_images/11158618-52efd0 ...

  10. struct iphdr

    struct iphdr { #if defined(__LITTLE_ENDIAN_BITFIELD) __u8 ihl:, version:; #elif defined (__BIG_ENDIA ...