Minimum Height Trees
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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 undirectededges
(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 inedges
.
Example 1:
Given n = 4
, edges = [[1, 0], [1, 2], [1, 3]]
0 | 1 / \ 2 3
return [1]
Example 2:
Given n = 6
, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]
0 1 2 \ | / 3 | 4 | 5
return [3, 4]
Note:
(1) According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected byexactly one path. In other words, any connected graph without simple cycles is a tree.”
(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.
Solution:
public class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer> leaves = new ArrayList<Integer>();
if (edges==null || edges.length==0) {
leaves.add(n-1);
return leaves;
}
HashMap<Integer, ArrayList<Integer>> graph = new HashMap<Integer, ArrayList<Integer>>();
int[] indegree = new int[n];
for (int i=0; i<n; i++) {
graph.put(i, new ArrayList<Integer>());
}
//build the graph
for (int[] edge : edges) {
graph.get(edge[0]).add(edge[1]);
graph.get(edge[1]).add(edge[0]);
indegree[edge[0]]++;
indegree[edge[1]]++;
}
//find the leaves
for (int i=0; i<n; i++) {
if (indegree[i] == 1) {
leaves.add(i);
}
}
//topological sort until n<=2
while (n > 2) {
List<Integer> newLeaf = new ArrayList<Integer>();
for (Integer leaf : leaves) {
List<Integer> neighbors = graph.get(leaf);
for (Integer neighbor : neighbors) {
indegree[neighbor]--;
graph.get(neighbor).remove(leaf);
if (indegree[neighbor] == 1)
newLeaf.add(neighbor);
}
//delete leaf from graph
n--;
}
leaves = newLeaf;
}
return leaves;
}
}
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