我已经有了在python中实现红黑树的代码。但是，当我运行我的代码时，它似乎只是插入值，就好像它是一个普通的BST，然后将所有内部顶点着色为红色。这是家庭作业的一项实践任务，但我陷入困境，无处可去。
这是我的实施：
class BinaryTreeVertex(object):
def __init__(self, data=None, isALeaf=True, colour='r', left=None, right=None):
# Assume user will pass a left & right ONLY IF isALeaf=False
self.data = data
self.isALeaf = isALeaf
self.colour = colour
self.left = left
self.right = right
class BinarySearchTree(object):
def __init__(self, root=None):
self.root = BinaryTreeVertex()
def insert(self, value):
self.root = recInsert(self.root, value)
self.root.colour = 'b'
def searchPath(self, val):
path = []
node = self.root
while node is not None:
path.append(node.data)
if val < node.data:
node = node.left
elif val > node.data:
node = node.right
else:
node = None
return path
def totalDepth(self):
if self.root.isALeaf:
return 0
else:
return recTotalDepth(self.root)
# returns the depth of a given node
# plus it's children's depths
def recTotalDepth(node, currDepth=0):
if node.isALeaf:
return 0
else:
leftDepth = 0
rightDepth = 0
if node.left is not None:
leftDepth = recTotalDepth(node.left, currDepth+1)
if node.right is not None:
rightDepth = recTotalDepth(node.right, currDepth+1)
return leftDepth + currDepth + rightDepth
# print a sideways representation of the BinarySearchTree
# Up = right, down = left
def printTree(node, indent=0):
if node.isALeaf:
return
else:
if node.right is not None:
printTree(node.right, indent+4)
print(" "*indent + str(node.data) + node.colour)
if node.left is not None:
printTree(node.left, indent+4)
# Insert a value into the binary search tree
def recInsert(node, value):
if node.isALeaf:
# Set the data to value, the colour to red, give the vertex two
# empty leaves coloured black
return BinaryTreeVertex(value, False, 'r', BinaryTreeVertex(None, True, 'b'), BinaryTreeVertex(None, True, 'b'))
elif value > node.data:
node.right = recInsert(node.right, value)
# check for balance
if node.colour == 'r':
# no balance check @ red vertices
return node
else: # node.colour is black
# Grandparents/Parents will handle balance of red vertices
# We must only perform balances at black vertices
if node.right.colour == 'r':
# right child is red, possibility for red-red conflict
if node.right.right.colour == 'r':
# red-red conflict on the right-right children
return rightRightFix(node)
elif node.right.left.colour == 'r':
# red-red conflict on the right-left children
return rightLeftFix(node)
else:
# no red-red conflicts
return node
else:
# right child is black, no fixing needed
return node
else: # value < self.data
node.left = recInsert(node.left, value)
# check for balance
if node.colour == 'r':
# no balance checks @ red vertices
return node
else: # node.colour == 'b'
# Grandparents/Parents will handle balance of red vertices
# We must only perform balances at black vertices
if node.left.colour == 'r':
# left child is red, possibility for red-red conflict
if node.left.left.colour == 'r':
# red-red conflict on the left-left children
return leftLeftFix(node)
elif node.left.right.colour == 'r':
# red-red conflict on the left-right children
return leftRightFix(node)
else:
# no red-red conflicts
return node
else:
# left child is black, no fixing needed
return node
def rightRightFix(node):
# red-red conflict on right-right children
child = node.right
sib = node.left
if sib.colour == 'r':
# no need for rotation, just recolour
child.colour == 'b'
sib.colour == 'b'
node.colour == 'r'
return node
else:
# sib's colour is black, single rot
# fix pointers first
node.right = child.left
child.left = node
# fix colours
child.colour = 'b'
node.colour = 'r'
# make the new root, which is now child
return child
def rightLeftFix(node):
# red-red conflict on right-left children
child = node.right
sib = node.left
if sib.colour == 'r':
# no need for rotation, just recolour
child.colour == 'b'
sib.colour == 'b'
node.colour == 'r'
return node
else:
# sib's colour is black, double rot
# fix the pointers
grandchild = child.left
child.left = grandchild.right
node.right = grandchild.left
grandchild.left = node
grandchild.right = child
# fix the colours
grandchild.colour == 'b'
node.colour = 'r'
# return the new root, which is now grandchild
return grandchild
def leftLeftFix(node):
# red-red conflict on right-right children
child = node.left
sib = node.right
if sib.colour == 'r':
# no need for rotation, just recolour
child.colour == 'b'
sib.colour == 'b'
node.colour == 'r'
return node
else:
# sib's colour is black, single rot
# fix pointers first
node.left = child.right
child.right = node
# fix colours
child.colour = 'b'
node.colour = 'r'
# make the new root, which is now child
return child
def leftRightFix(node):
# red-red conflict on left-right children
child = node.left
sib = node.right
if sib.colour == 'r':
# no need for rotation, just recolour
child.colour == 'b'
sib.colour == 'b'
node.colour == 'r'
return node
else:
# sib's colour is black, double rot
# fix the pointers
grandchild = child.right
child.right = grandchild.left
node.left = grandchild.right
grandchild.right = node
grandchild.left = child
# fix the colours
grandchild.colour == 'b'
node.colour = 'r'
# return the new root, which is now grandchild
return grandchild
def main():
myTree = BinarySearchTree()
myList = [13,42,3,6,23,32,72,90,1,10,26,85,88,97,73,80,35,36,88,34,12,92,100,143,123,124,125,126,127,128]
for v in myList:
myTree.insert(v)
printTree(myTree.root)
print(myTree.searchPath(12))
print(myTree.totalDepth())
myTree2 = BinarySearchTree()
myList2 = [6, 10, 20, 8, 3]
for v in myList2:
myTree2.insert(v)
printTree(myTree2.root)
print(myTree2.searchPath(6))
print(myTree2.totalDepth())
main()
答案 0 :(得分：1)
所提供来源中的主要问题是解释为什么您的算法 “实际上不会重新找回我的顶点” 是由于分配不当造成的。
在所有函数rightRightFix()，rightLeftFix()，leftLeftFix()和leftRightFix()中，神秘地将某些颜色分配（child.colour = 'b'）编码为比较（{{1 }}）。
解决方案 - 只需更换14个错误的复制/粘贴分配错误。
在纠正之后，您的红黑树算法运行良好。
次要错误 - 在child.colour == 'b'找不到值时。
当搜索的值不存在时，您会得到：
BinarySearchTree::searchPath()
这是由于使用了不初始化Leaf节点。
TypeError: unorderable types: int() < NoneType()在这种情况下，返回的搜索路径将以'None'=＆gt;结束。找不到
输入：
def searchPath(self, val): path = [] node = self.root while node is not None: path.append(node.data) # break the while when node is the leave if (node.isALeaf == True): break if val < node.data: node = node.left elif val > node.data: node = node.right else: node = None return path输出：
print(myTree.searchPath(102))