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Question
Formatted question description: https://leetcode.ca/all/310.html
Level
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.
- Start storing all nodes (leaf nodes) with only one connected edge into a queue
- 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
- If the node becomes a leaf node after deletion, add it to the queue and delete it in the next round
- Stop when the number of nodes is
less than or equal to 2
. At this time, theremaining one or two
nodes are the root nodes of the minimum height tree we require
Code
-
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) { graph.add(new HashSet<>()); } for (int[] edge : edges) { graph.get(edge[0]).add(edge[1]); graph.get(edge[1]).add(edge[0]); } // en-queue for leaf node List<Integer> leaves = new ArrayList<>(); for (int i = 0; i < n; ++i) { if (graph.get(i).size() == 1) { leaves.add(i); } } // 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) { newLeaves.add(newLeaf); } } n -= leaves.size(); leaves = newLeaves; } return leaves; } } } ############ class Solution { public List<Integer> findMinHeightTrees(int n, int[][] edges) { if (n == 1) { return Collections.singletonList(0); } List<Integer>[] g = new List[n]; Arrays.setAll(g, k -> new ArrayList<>()); int[] degree = new int[n]; for (int[] e : edges) { int a = e[0], b = e[1]; g[a].add(b); g[b].add(a); ++degree[a]; ++degree[b]; } Queue<Integer> q = new LinkedList<>(); for (int i = 0; i < n; ++i) { if (degree[i] == 1) { q.offer(i); } } List<Integer> ans = new ArrayList<>(); while (!q.isEmpty()) { ans.clear(); for (int i = q.size(); i > 0; --i) { int a = q.poll(); ans.add(a); for (int b : g[a]) { if (--degree[b] == 1) { q.offer(b); } } } } return ans; } }
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// 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; } };
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from typing import List, Set from collections import defaultdict class Solution: def findMinHeightTrees(self, n: int, edges: List[List[int]]) -> List[int]: if n == 1: return [0] graph = defaultdict(set) for a, b in edges: graph[a].add(b) graph[b].add(a) # just check neighbours[] size, not indgree[] for each node leaves = [i for i in range(n) if len(graph[i]) == 1] while n > 2: n -= len(leaves) new_leaves = [] for leaf in leaves: neighbor = graph[leaf].pop() # should be only one in hashset, because it's a leaf node graph[neighbor].remove(leaf) if len(graph[neighbor]) == 1: new_leaves.append(neighbor) leaves = new_leaves return leaves ############### class Solution: def findMinHeightTrees(self, n: int, edges: List[List[int]]) -> List[int]: if n == 1: return [0] g = defaultdict(list) degree = [0] * n for a, b in edges: g[a].append(b) g[b].append(a) degree[a] += 1 # not needed, just len(g[a]) is good enough, will update it in next round updating degree[b] += 1 q = deque() for i in range(n): if degree[i] == 1: q.append(i) ans = [] while q: n = len(q) ans.clear() for _ in range(n): a = q.popleft() ans.append(a) for b in g[a]: degree[b] -= 1 if degree[b] == 1: # final round only 2 left (a,b), then degree[b] here will be 0, and no more node enqueue q.append(b) 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]]
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func findMinHeightTrees(n int, edges [][]int) []int { if n == 1 { return []int{0} } g := make([][]int, n) degree := make([]int, n) for _, e := range edges { a, b := e[0], e[1] g[a] = append(g[a], b) g[b] = append(g[b], a) degree[a]++ degree[b]++ } var q []int for i := 0; i < n; i++ { if degree[i] == 1 { q = append(q, i) } } var ans []int for len(q) > 0 { ans = []int{} for i := len(q); i > 0; i-- { a := q[0] q = q[1:] ans = append(ans, a) for _, b := range g[a] { degree[b]-- if degree[b] == 1 { q = append(q, b) } } } } return ans }