310 Minimum Height Trees

310. Minimum Height Trees

1. Question

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 containsnnodes which are labeled from0ton - 1. You will be given the numbernand a list of undirectededges(each edge is a pair of labels).

You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges.

Example 1:

Givenn = 4,edges = [[1, 0], [1, 2], [1, 3]]

        0
        |
        1
       / \
      2   3

return[1]

Example 2:

Givenn = 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 byexactlyone 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.

2. Implementation

(1) BFS

class Solution {
    public List<Integer> findMinHeightTrees(int n, int[][] edges) {
        List<Integer> res = new ArrayList<>();

        if (n == 1) {
            res.add(0);
            return res;
        }

        List<Set<Integer>> adjList = new ArrayList<>();

        for (int i = 0; i < n; i++) {
            adjList.add(new HashSet<>());
        }

        for (int[] edge : edges) {
            adjList.get(edge[0]).add(edge[1]);
            adjList.get(edge[1]).add(edge[0]);
        }

        List<Integer> leaves = new ArrayList<>();
        List<Integer> newLeaves = null;

        for (int i = 0; i < n; i++) {
            if (adjList.get(i).size() == 1) {
                leaves.add(i);
            }
        }

        while (n > 2) {
            n -= leaves.size();
            newLeaves = new ArrayList<>();

            for (int node : leaves) {
                for (int parent : adjList.get(node)) {
                    adjList.get(parent).remove(node);
                    if (adjList.get(parent).size() == 1) {
                        newLeaves.add(parent);
                    }
                }
            }
            leaves = newLeaves;
        }
        return leaves;
    }
}

3. Time & Space Complexity

BFS: 时间复杂度O(n)， 空间复杂度O(n)