Minimum Height Trees

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:

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]

Hint:

How many MHTs can a graph have at most? Note:

(1) 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.”

(2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf.

Solution

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_list, unvisited = [], set()
        for i in range(n):
            if len(neighbors[i]) == 1: ## this is a leaf
                pre_list.append(i)
            unvisited.add(i)

        while len(unvisited) > 2:
            cur_list = []
            for u in pre_list:
                unvisited.remove(u)
                for v in neighbors[u]:
                    if v in unvisited:
                        neighbors[v].remove(u)
                        if len(neighbors[v]) == 1:
                            cur_list.append(v)
            pre_list = cur_list

        return list(unvisited)