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minimum-height-trees.py
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minimum-height-trees.py
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# Time: O(n)
# Space: O(n)
# 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.
import collections
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_level, unvisited = [], set()
for i in xrange(n):
if len(neighbors[i]) == 1: # A leaf.
pre_level.append(i)
unvisited.add(i)
# A graph can have 2 MHTs at most.
# BFS from the leaves until the number
# of the unvisited nodes is less than 3.
while len(unvisited) > 2:
cur_level = []
for u in pre_level:
unvisited.remove(u)
for v in neighbors[u]:
if v in unvisited:
neighbors[v].remove(u)
if len(neighbors[v]) == 1:
cur_level.append(v)
pre_level = cur_level
return list(unvisited)