# Question

Formatted question description: https://leetcode.ca/all/310.html

Medium

## Description

For an 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.

# Algorithm

Any connected graph without simple cycles is a tree.

1. Start storing all nodes (leaf nodes) with only one connected edge into a queue
2. Then we traverse each leaf node, find the node connected to it through the graph, and delete the leaf node from the set of connected nodes
3. If the node becomes a leaf node after deletion, add it to the queue and delete it in the next round
4. Stop when the number of nodes is less than or equal to 2. At this time, the remaining one or two nodes are the root nodes of the minimum height tree we require

# Code

Java

• import java.util.*;

public class Minimum_Height_Trees {

public class Solution {

// @note: bfs to find min steps
public List<Integer> findMinHeightTrees(int n, int[][] edges) {

if (n == 1) {
return Collections.singletonList(0);
}

List<Set<Integer>> graph = new ArrayList<>(n);

// build graph
for (int i = 0; i < n; ++i) {
}
for (int[] edge : edges) {
}

// en-queue for leaf node
List<Integer> leaves = new ArrayList<>();
for (int i = 0; i < n; ++i) {
if (graph.get(i).size() == 1) {
}
}

// batch removing leaf nodes
while (n > 2) { // if n>=3 then can be further reduced
List<Integer> newLeaves = new ArrayList<>();
for (int oldLeaf: leaves) {
// leaves.remove(oldLeaf); // @note: arraylist remove by object
int newLeaf = graph.get(oldLeaf).iterator().next(); // should be only one in hashset, because it's a leaf node
graph.get(newLeaf).remove(oldLeaf); // @note: hashset remove by object
if (graph.get(newLeaf).size() == 1) {
}
}

n -= leaves.size();
leaves = newLeaves;
}

return leaves;
}
}
}

• // OJ: https://leetcode.com/problems/minimum-height-trees
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
vector<int> findMinHeightTrees(int n, vector<vector<int>>& E) {
if (n == 1) return { 0 };
vector<int> degree(n), ans;
vector<vector<int>> G(n);
for (auto &e : E) {
int u = e[0], v = e[1];
degree[u]++;
degree[v]++;
G[u].push_back(v);
G[v].push_back(u);
}
queue<int> q;
for (int i = 0; i < n; ++i) {
if (degree[i] == 1) q.push(i);
}
while (n > 2) {
int cnt = q.size();
n -= cnt;
while (cnt--) {
int u = q.front();
q.pop();
for (int v : G[u]) {
if (--degree[v] == 1) q.push(v);
}
}
}
while (q.size()) {
ans.push_back(q.front());
q.pop();
}
return ans;
}
};

• from collections import deque

class Solution(object):
def findMinHeightTrees(self, n, edges):
"""
:type n: int
:type edges: List[List[int]]
:rtype: List[int]
"""
if len(edges) == 0:
if n > 0:
return [0]
return []

def bfs(graph, start):
queue = deque([(start, None)])
level = 0
maxLevel = -1
farthest = None
while queue:
level += 1
for i in range(0, len(queue)):
label, parent = queue.popleft()
for child in graph.get(label, []):
if child != parent:
queue.append((child, label))
if level > maxLevel:
maxLevel = level
farthest = child
return farthest

def dfs(graph, start, end, visited, path, res):
if start == end:
res.append(path + [])
return True
visited[start] = 1
for child in graph.get(start, []):
if visited[child] == 0:
path.append(child)
if dfs(graph, child, end, visited, path, res):
return True
path.pop()

graph = {}
for edge in edges:
graph[edge[0]] = graph.get(edge[0], []) + [edge[1]]
graph[edge[1]] = graph.get(edge[1], []) + [edge[0]]

start = bfs(graph, edges[0][0])
end = bfs(graph, start)
res, visited = [], [0 for i in range(0, n)]
dfs(graph, start, end, visited, [start], res)
if not res:
return []
path = res[0]
if len(path) % 2 == 0:
return [path[len(path) / 2 - 1], path[len(path) / 2]]
else:
return [path[len(path) / 2]]