def top_view(self, aggregate_fn=_default_printfn, **kwargs):
if not self.root:
return
q = Queue()
min_width = 0
max_width = 0
q.enqueue((0, self.root))
# Top-view begins with the root node
aggregate_fn(kwargs, self.root.value)
while q.length() != 0:
width, node = q.dequeue()
if width < min_width:
min_width = width
aggregate_fn(kwargs, node.value)
elif width > max_width:
max_width = width
aggregate_fn(kwargs, node.value)
# NOTE: width 0, is root, would already be filled in at root,
# and in a top-view is not going to replaced
q.enqueue((width - 1, node.left)) if node.left else None
q.enqueue((width + 1, node.right)) if node.right else None
def width(self):
if not self.root:
return 0
q = Queue()
q.enqueue((0, self.root))
curr_level = 0
curr_level_nodes = 0
max_width = 0
while q.length() != 0:
level, node = q.dequeue()
q.enqueue((level + 1, node.left)) if node.left else None
q.enqueue((level + 1, node.right)) if node.right else None
# Start of a new level
# Check if number of nodes at previous level > mqx nodes at any level
if level != curr_level:
if curr_level_nodes > max_width:
max_width = curr_level_nodes
# start counting nodes for the current level
curr_level_nodes = 0
curr_level_nodes += 1
curr_level = level
# Check last level
if curr_level_nodes > max_width:
max_width = curr_level_nodes
return max_width
def all_shortest_paths_by_length(self, v1, v2, aggregate_fn=None, *args, **kwargs): q = Queue() q.enqueue((v1, [], 0)) # Maximum number of vertices wont exceed the number of vertices # if there are cycles, they'd not be included more than once in the path shortest_path_level = self.vertices while q: curr_vertex, curr_prefix, level = q.dequeue() # Found all the shortest paths if level > shortest_path_level: return # found destination vertex, # record this path as one of the paths if curr_vertex == v2: path = curr_prefix + [v2] shortest_path_level = level aggregate_fn(path, *args, **kwargs) for v,_ in self._adjlists[curr_vertex]: # visited[] traking doesn't yield well to BFS # when extracting all paths # Instead just check if we don't add a vertex already # in the current path so we don't loop endlessly # if there's a cycle / its an undirected graph if v not in curr_prefix: q.enqueue((v, curr_prefix + [curr_vertex], level+1))
class QueueMin(object):
def __init__(self):
self.q = Queue()
self.min = None
# All queue items are always in Q,
def __len__(self):
return self.q.size
def __repr__(self):
return "{%r}/Min: %r" % (self.q, self.min)
def __str__(self):
return str(self.q)
def minimum(self):
return self.min
# Add to Q and return
def enqueue(self, x):
self.q.enqueue(x)
# update min
# if min is None => this is the first item to be enqueued
self.min = min(x, self.min) if self.min else x
# Dequeue() from Q
# if popped item was minima, Re-enqueue all items while calculating new minima
def dequeue(self):
# Helper function to re-enqueue all items from Queue, Q back into itself
# while updating new minima
def reenqueue(q):
minima = q.front()
for i in range(len(q)):
x = q.dequeue()
q.enqueue(x)
minima = min(x, minima)
return minima
# SLL raises UnderFlowError if Q is empty
item = self.q.dequeue()
# We just dequeued the last item from the queue
# restore min to be None
if not self.q:
self.min = None
elif item == self.min:
# Update min if item == minima
self.min = reenqueue(self.q)
return item
# Return the front of Q
def front(self):
return self.q.front()
# Return the back of Q
def back(self):
return self.q.back()
def levelorder_traversal(self, aggregate_fn=_default_printfn, **kwargs):
if not self.root:
return
q = Queue()
q.enqueue(self.root)
while q.length() != 0:
tmp = q.dequeue()
aggregate_fn(kwargs, tmp.value)
q.enqueue(tmp.left) if tmp.left else None
q.enqueue(tmp.right) if tmp.right else None
class Stack(object):
def __init__(self):
self.q1 = Queue()
self.q2 = Queue()
# peek what's at the top of the stack
# New items are always added to Q1's end
# Stack's top would be at the back of Q1
def top(self):
return self.q1.back()
def __len__(self):
return self.q1.size
# Just add to Q1 and return
def push(self, x):
self.q1.enqueue(x)
# Move (n-1) items from q1 to q2
# dequeue remaining item from q1 to return
# swap q1, q2
def pop(self):
# Helper function to move 'n' items from queue a to queue b
def move(a, b, n):
for i in range(n):
b.enqueue(a.dequeue())
# Move (n-1) items from q1 to q2
move(self.q1, self.q2, self.q1.size-1)
# Swap Q1, Q2
self.q1, self.q2 = self.q2, self.q1
# Dequeue and return the last item from what was q1 earlier, and now q2
return self.q2.dequeue()
def __str__(self):
slist = []
for x in self.q1:
slist.insert(0, str(x))
return '[%d]: ' %(self.__len__()) + ' '.join(slist)
def __repr__(self):
return "{%r , %r}" %(self.q1, self.q2)
def bfs(adjacency_lists, startvertex, visited, aggregate_fn, *args, **kwargs): neighbors = Queue() neighbors.enqueue(startvertex) visited[startvertex] = True while neighbors: v = neighbors.dequeue() aggregate_fn(v, *args, **kwargs) # enqueue all vertices that are neighbors of v for n,_ in adjacency_lists[v]: if not visited[n]: visited[n] = True neighbors.enqueue(n)
def bfs_i(adjacency_matrix, startvertex, visited, aggregate_fn, *args, **kwargs): neighbors = Queue() neighbors.enqueue(startvertex) while neighbors: v = neighbors.dequeue() if not visited[v]: visited[v] = True aggregate_fn(v, *args, **kwargs) # enqueue all vertices that are neighbors of v for n in xrange(len(visited)): if adjacency_matrix[v][n] is not None and not visited[n]: neighbors.enqueue(n)
def bfs_i(adjacency_lists, startvertex, visited, aggregate_fn, *args, **kwargs): neighbors = Queue() neighbors.enqueue(startvertex) while neighbors: v = neighbors.dequeue() if not visited[v]: visited[v] = True aggregate_fn(v, *args, **kwargs) # enqueue all vertices that are neighbors of v for n,_ in adjacency_lists[v]: if not visited[n]: neighbors.enqueue(n)
def bfs(adjacency_matrix, startvertex, visited, aggregate_fn, *args, **kwargs): neighbors = Queue() visited[startvertex] = True neighbors.enqueue(startvertex) while neighbors: v = neighbors.dequeue() aggregate_fn(v, *args, **kwargs) # enqueue all vertices that are neighbors of v for n in xrange(len(visited)): if adjacency_matrix[v][n] is not None and not visited[n]: visited[n] = True neighbors.enqueue(n)
class Stack(object): def __init__(self): self.q = Queue() # peek what's at the top of the stack # New items are always added to Q's end # Stack's top would be at the back of Q def top(self): return self.q.back() def __len__(self): return self.q.size # Just add to Q and return def push(self, x): self.q.enqueue(x) # re-enqueue (n-1) items from Q, and re-enqueue back to Q def pop(self): # Helper function to reenqueue 'n' items from queue a back into it def reenqueue(a, n): for i in range(n): a.enqueue(a.dequeue()) # reenqueue (n-1) items from q reenqueue(self.q, self.q.size-1) # Dequeue from the front of q return self.q.dequeue() # Actual stack order would be the q in reverse order def __str__(self): slist = [] for x in self.q: slist.insert(0, str(x)) return '[%d]: ' %(self.__len__()) + ' '.join(slist) def __repr__(self): return "%r" %(self.q)
def top_view_LR(self, aggregate_fn=_default_printfn, **kwargs):
if not self.root:
return
# Queue to help with level-order traversal
q = Queue()
# Store L/R width of a node and its value if it is visible in the top-view, in a sorted list
# ordered by the width
# at the end of the level-order traversal the list would contain
# [(-max_left_width, node value), ...., (0, root.value), ... (max_right_width, node value)]
slist = SLL()
# pairs a(w1, node_val1) vs b(w2, node_val2) are valued against each other
# depending on w1 vs w2 (where w1 and w2 are widths, so slist is kept sorted by the nodes' L/R widths)
pair_comparator = lambda (w1, node_val1), (w2, node_val2): cmp(w1, w2)
min_width = 0
max_width = 0
q.enqueue((0, self.root))
# Top-view begins with the root node
slist.place((0, self.root.value))
while q.length() != 0:
width, node = q.dequeue()
if width < min_width:
min_width = width
slist.place((width, node.value))
elif width > max_width:
max_width = width
slist.place((width, node.value))
# NOTE: width 0, is root, would already be filled in at root,
# and in a top-view is not going to replaced
q.enqueue((width - 1, node.left)) if node.left else None
q.enqueue((width + 1, node.right)) if node.right else None
# At the end of the level-order traversal,
# slist is ordered by width, so
# (-max_left_width, ..., 0, ..., max_right_width)
# Just retrieve the SLL L-R for the top view (L-R)
while slist.size != 0:
width, item = slist.pop_front()
aggregate_fn(kwargs, item)
class Stack(object):
def __init__(self):
self.q1 = Queue()
self.q2 = Queue()
# peek what's at the top of the stack
# New items are always added to Q1's end
# Stack's top would be at the back of Q1
def top(self):
return self.q1.back()
def __len__(self):
return self.q1.size
# Just add to Q1 and return
def push(self, x):
self.q1.enqueue(x)
# Move (n-1) items from q1 to q2
# dequeue remaining item from q1 to return
# swap q1, q2
def pop(self):
# Helper function to move 'n' items from queue a to queue b
def move(a, b, n):
for i in range(n):
b.enqueue(a.dequeue())
# Move (n-1) items from q1 to q2
move(self.q1, self.q2, self.q1.size - 1)
# Swap Q1, Q2
self.q1, self.q2 = self.q2, self.q1
# Dequeue and return the last item from what was q1 earlier, and now q2
return self.q2.dequeue()
def __str__(self):
slist = []
for x in self.q1:
slist.insert(0, str(x))
return '[%d]: ' % (self.__len__()) + ' '.join(slist)
def __repr__(self):
return "{%r , %r}" % (self.q1, self.q2)
class Stack(object):
def __init__(self):
self.q1 = Queue()
self.q2 = Queue()
# peek what's at the top of the stack
# Stack-order is preserved in Q1
# Stack's top would be at the front of Q1
def top(self):
return self.q1.front()
def __len__(self):
return self.q1.size
# Add x to Q2
# Move everything from Q1 to Q2, so Q2 preserves stack-order
# swap Q1, Q2
def push(self, x):
# Helper function to move all items from queue a to queue b
def move(a, b):
while a:
b.enqueue(a.dequeue())
self.q2.enqueue(x)
# Move items from q1 to q2
move(self.q1, self.q2)
# Swap Q1, Q2
self.q1, self.q2 = self.q2, self.q1
# Q1 has items in stack-order
# Just dequeue from Q1 and return
def pop(self):
return self.q1.dequeue()
def __str__(self):
return str(self.q1)
def __repr__(self):
return "{%r , %r}" %(self.q1, self.q2)
def left_view(self, aggregate_fn=_default_printfn, **kwargs):
if not self.root:
return
q = Queue()
levels_done = None
q.enqueue((0, self.root))
while q.length() != 0:
curr_level, node = q.dequeue()
# Nothing has been printed in this level so far
# NOTE: (None < 0) in python
if curr_level > levels_done:
aggregate_fn(kwargs, node.value)
levels_done = curr_level
q.enqueue((curr_level + 1, node.left)) if node.left else None
q.enqueue((curr_level + 1, node.right)) if node.right else None
class Stack(object): def __init__(self): self.q = Queue() # peek what's at the top of the stack # Stack-order is preserved in Q # Stack's top would be at the front of Q def top(self): return self.q.front() def __len__(self): return self.q.size # Enqueue x into Q # re-enqueue (n-1) items from Q, and re-enqueue back to Q def push(self, x): # Helper function to reenqueue 'n' items from queue a back into it def reenqueue(a, n): for i in range(n): a.enqueue(a.dequeue()) self.q.enqueue(x) # reenqueue (n-1) items from q reenqueue(self.q, self.q.size-1) # Just dequeue from Q as it preserves stack-order def pop(self): # Dequeue from the front of q return self.q.dequeue() # Q preserves stack-order, Just return Q as-is def __str__(self): return str(self.q) def __repr__(self): return "%r" %(self.q)
def span(self):
if not self.root:
return 0
q = Queue()
min_width = 0
max_width = 0
q.enqueue((0, self.root))
while q.length() != 0:
width, node = q.dequeue()
if width < min_width:
min_width = width
elif width > max_width:
max_width = width
q.enqueue((width - 1, node.left)) if node.left else None
q.enqueue((width + 1, node.right)) if node.right else None
return max_width - min_width
def bottom_view(self, aggregate_fn=_default_printfn, **kwargs):
if not self.root:
return
q = Queue()
# A hash table that stores a map of L/R width to a node's item
# We traverse the tree in level-order, and keep replacing slist[width] if we find a new node with the same width
# At the end of the traversal, every slist[width] from {-max_left_width, ..., 0, ..., max_right_width} will contain the
# node items that would be visible from a bottom view of the binary tree
slist = {}
min_width = 0
max_width = 0
q.enqueue((0, self.root))
# Star with placing root at width 0
slist[0] = self.root.value
while q.length() != 0:
width, node = q.dequeue()
# As we traverse level-by-level, keep replacing
# item at L/R width
slist[width] = node.value
# Track max_left_width, max_right_width
# so we don't need to sort the hash table slist,
# Just retrieve slist[{-max_left_width, ..., 0, ..., max_right_width}]
if width < min_width:
min_width = width
elif width > max_width:
max_width = width
q.enqueue((width - 1, node.left)) if node.left else None
q.enqueue((width + 1, node.right)) if node.right else None
# At the end of the level-order traversal,
# Just 'aggregate' slist[{-max_left_width, ..., 0, ..., max_right_width}]
for i in range(min_width, max_width + 1):
aggregate_fn(kwargs, slist[i])
class Stack(object):
def __init__(self):
self.q = Queue()
# peek what's at the top of the stack
# New items are always added to Q's end
# Stack's top would be at the back of Q
def top(self):
return self.q.back()
def __len__(self):
return self.q.size
# Just add to Q and return
def push(self, x):
self.q.enqueue(x)
# re-enqueue (n-1) items from Q, and re-enqueue back to Q
def pop(self):
# Helper function to reenqueue 'n' items from queue a back into it
def reenqueue(a, n):
for i in range(n):
a.enqueue(a.dequeue())
# reenqueue (n-1) items from q
reenqueue(self.q, self.q.size - 1)
# Dequeue from the front of q
return self.q.dequeue()
# Actual stack order would be the q in reverse order
def __str__(self):
slist = []
for x in self.q:
slist.insert(0, str(x))
return '[%d]: ' % (self.__len__()) + ' '.join(slist)
def __repr__(self):
return "%r" % (self.q)
class Stack(object):
def __init__(self):
self.q1 = Queue()
self.q2 = Queue()
# peek what's at the top of the stack
# Stack-order is preserved in Q1
# Stack's top would be at the front of Q1
def top(self):
return self.q1.front()
def __len__(self):
return self.q1.size
# Add x to Q2
# Move everything from Q1 to Q2, so Q2 preserves stack-order
# swap Q1, Q2
def push(self, x):
# Helper function to move all items from queue a to queue b
def move(a, b):
while a:
b.enqueue(a.dequeue())
self.q2.enqueue(x)
# Move items from q1 to q2
move(self.q1, self.q2)
# Swap Q1, Q2
self.q1, self.q2 = self.q2, self.q1
# Q1 has items in stack-order
# Just dequeue from Q1 and return
def pop(self):
return self.q1.dequeue()
def __str__(self):
return str(self.q1)
def __repr__(self):
return "{%r , %r}" % (self.q1, self.q2)
def paths_2(self, v1, v2, aggregate_fn=None, *args, **kwargs): if not aggregate_fn: aggregate_fn = Graph._default_printfn q = Queue() q.enqueue((v1, [])) while q: curr_vertex, curr_prefix = q.dequeue() # found destination vertex, # record this path as one of the paths if curr_vertex == v2: aggregate_fn(curr_prefix + [v2], *args, **kwargs) for v,_ in self._adjlists[curr_vertex]: # visited[] traking doesn't yield well to BFS # when extracting all paths # Instead just check if we don't add a vertex already # in the current path so we don't loop endlessly # if there's a cycle / its an undirected graph if v not in curr_prefix: q.enqueue((v, curr_prefix + [curr_vertex]))
def shortest_path_by_length(self, v1, v2):
q = Queue()
retrace = {}
q.enqueue(v1)
retrace[v1] = None
def retrace_path(retrace):
path = []
v = v2
while v is not None:
path.insert(0, v)
v = retrace[v]
return path
# Start a BFS traversal
while q:
curr_vertex = q.dequeue()
# found destination vertex,
# retrace from last vertex using the mapping all the way to v1
if curr_vertex == v2:
return retrace_path(retrace)
for v,_ in self._adjlists[curr_vertex]:
# visited[] traking doesn't yield well to BFS
# when extracting all paths
# Instead just check if we don't add a vertex already
# in the current path so we don't loop endlessly
# if there's a cycle / its an undirected graph
if not retrace.has_key(v):
q.enqueue(v)
retrace[v] = curr_vertex
return []
def right_view(self, aggregate_fn=_default_printfn, **kwargs):
if not self.root:
return
q = Queue()
q.enqueue((0, self.root))
while q.length() != 0:
curr_level, node = q.dequeue()
q.enqueue((curr_level + 1, node.left)) if node.left else None
q.enqueue((curr_level + 1, node.right)) if node.right else None
# peek next node in the queue to see if this is the last node in the current level
# if yes, print it
try:
(level, next_entry) = q.front()
except UnderFlowError:
(level, next_entry) = (None, None)
# Queue is either empty, in which case this is the rightmost node in the last level
# *OR*, next entry in the queue is for the level after this one, so this is the rightmost in the current level
# In both cases, current node would be visible in a right-view.
if (not next_entry) or (level > curr_level):
aggregate_fn(kwargs, node.value)
class Stack(object):
def __init__(self):
self.q = Queue()
# peek what's at the top of the stack
# Stack-order is preserved in Q
# Stack's top would be at the front of Q
def top(self):
return self.q.front()
def __len__(self):
return self.q.size
# Enqueue x into Q
# re-enqueue (n-1) items from Q, and re-enqueue back to Q
def push(self, x):
# Helper function to reenqueue 'n' items from queue a back into it
def reenqueue(a, n):
for i in range(n):
a.enqueue(a.dequeue())
self.q.enqueue(x)
# reenqueue (n-1) items from q
reenqueue(self.q, self.q.size - 1)
# Just dequeue from Q as it preserves stack-order
def pop(self):
# Dequeue from the front of q
return self.q.dequeue()
# Q preserves stack-order, Just return Q as-is
def __str__(self):
return str(self.q)
def __repr__(self):
return "%r" % (self.q)
class QueueMin(object):
def __init__(self):
self.q = Queue()
self.min = None
# All queue items are always in Q,
def __len__(self):
return self.q.size
def __repr__(self):
return "{%r}/Min: %r" %(self.q, self.min)
def __str__(self):
return str(self.q)
def minimum(self):
return self.min
# Add to Q and return
def enqueue(self, x):
self.q.enqueue(x)
# update min
# if min is None => this is the first item to be enqueued
self.min = min(x, self.min) if self.min else x
# Dequeue() from Q
# if popped item was minima, Re-enqueue all items while calculating new minima
def dequeue(self):
# Helper function to re-enqueue all items from Queue, Q back into itself
# while updating new minima
def reenqueue(q):
minima = q.front()
for i in range(len(q)):
x = q.dequeue()
q.enqueue(x)
minima = min(x, minima)
return minima
# SLL raises UnderFlowError if Q is empty
item = self.q.dequeue()
# We just dequeued the last item from the queue
# restore min to be None
if not self.q:
self.min = None
elif item == self.min:
# Update min if item == minima
self.min = reenqueue(self.q)
return item
# Return the front of Q
def front(self):
return self.q.front()
# Return the back of Q
def back(self):
return self.q.back()
class QueueMin(object):
def __init__(self):
self.q1 = Queue()
self.dq2 = Deque()
self.min = None
# All queue items are always in Q1,
# DQ2 is merely used to store relative order of minimas
def __len__(self):
return self.q1.size
def __repr__(self):
return "{%r , %r}/Min: %r" %(self.q1, self.dq2, self.min)
def __str__(self):
return str(self.q1)
def minimum(self):
return self.min
# Add item to Q1
def enqueue(self, x):
self.q1.enqueue(x)
# update min
# if min is None => this is the first item to be enqueued
self.min = min(x, self.min) if self.min else x
# Update DQ2 to maintain the order of minimas,
# so when a minima is removed, the next minima can be immediately retrieved
# remove everything from the back of DQ2 as long as it is > item x
while self.dq2 and self.dq2.back() > x:
self.dq2.pop_back()
# Add item x to DQ2
self.dq2.push_back(x)
# Dequeue() from Q1
# DQ2 stores relative order of minimas for the current queue items
# if popped item was minima, update minima to DQ2 new head()
def dequeue(self):
# SLL raises UnderFlowError if Q2 is empty
item = self.q1.dequeue()
# item removed was the current minima, remove it from DQ2 as well
# item would be found in the head of DQ2
if item == self.min:
self.dq2.pop_front()
# Also, update new min which would be the new head of DQ2 after removing the current minima
# NOTE: self.min should be None, if dq2 is empty - ie, we just removed the last item in the Queue
self.min = self.dq2.front() if self.dq2 else None
return item
# Return the front of Q1
def front(self):
return self.q1.front()
# Return the back of Q1
def back(self):
return self.q1.back()
class QueueMin(object):
def __init__(self):
self.q1 = Queue()
self.q2 = Queue()
self.min = None
# All queue items are always in Q1, Q2 is merely used to move queue elements and update minima
# after which Q2 is renamed Q1
def __len__(self):
return self.q1.size
def __repr__(self):
return "{%r , %r}/Min: %r" % (self.q1, self.q2, self.min)
def __str__(self):
return str(self.q1)
def minimum(self):
return self.min
# Add to Q1 and return
def enqueue(self, x):
self.q1.enqueue(x)
# update min
# if min is None => this is the first item to be enqueued
self.min = min(x, self.min) if self.min else x
# Dequeue() from Q2
# if popped item was minima, use Q2 to move all items while calculating new minima
def dequeue(self):
# Helper function to move all items from Queue, A to Queue, B
# while updating new minima
def move(a, b):
minima = a.front()
while a:
x = a.dequeue()
b.enqueue(x)
minima = min(x, minima)
return minima
# SLL raises UnderFlowError if Q2 is empty
item = self.q1.dequeue()
# We just dequeued the last item from the queue
# restore min to be None
if not self.q1:
self.min = None
elif item == self.min:
# Update min if item == minima
self.min = move(self.q1, self.q2)
# swap Q1, Q2
self.q1, self.q2 = self.q2, self.q1
return item
# Return the front of Q1
def front(self):
return self.q1.front()
# Return the back of Q1
def back(self):
return self.q1.back()
