LeetCode 310. Minimum Height Trees

Problem Statement

(Source) 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   3return [1]

Example 2:

Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]     0  1  2      \ | /        3        |        4        |        5return [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

Tags: Breadth-first Search, Graph.

class Solution(object):    def findMinHeightTrees(self, n, edges):        """        :type n: int        :type edges: List[List[int]]        :rtype: List[int]        """        edge = {i: set() for i in xrange(n)}        for e in edges:            u, v = e            edge[u].add(v)            edge[v].add(u)        res = [u for u in edge if len(edge[u]) <= 1]        while True:            temp = []            for u in res:                for v in edge[u]:                    edge[v].discard(u)                    if len(edge[v]) == 1:                        temp.append(v)            if not temp:                break            else:                for u in res:                    edge.pop(u)                res = temp        return res