我已经尝试解决在线挑战 9 天了。我有一个重复 100,000 次的在线插入删除，我需要找到其中的中位数。我尝试了两个堆，但实现随机删除不起作用。我现在开始使用二叉搜索树，因为它们似乎在 2 秒内在我的计算机上插入和删除了 100,000 个数据。我正在研究 python，我需要在 16 秒内运行。
我已经在线检查了解决方案，但完全不能满足我的需要。我发现在插入和删除时计算中位数可能是一个好策略。这我写了两个方法来获得一个节点的顺序后继或前驱。
问题是我认为是 - 在递归删除期间不正确地分配父指针。
我尝试了两种技术，但都效果不佳，如果我能得到一些帮助，我将不胜感激！
编码：
import sys
class BSTNode:
def __init__(self,x,parent):
self.data = x
self.left = None
self.right = None
self.count = 1
self.parent = parent
def delete(x,T):
if T is None:
print('Element Not Found')
elif x<T.data:
T.left = delete(x,T.left)
elif x>T.data:
T.right = delete(x,T.right)
elif T.left and T.right:
TempNode = findMin(T.right)
T.data = TempNode.data
T.right = delete(TempNode.data,T.right)
else:
if T.left is None:
T = T.right
elif T.right is None:
T = T.left
return T
def findMin(T):
if T.left:
return findMin(T.left)
else:
return T
def insert(x,T,parent=None):
if T is None:
T = BSTNode(x,parent)
elif x<T.data:
T.left = insert(x,T.left,T)
elif x>T.data:
T.right = insert(x,T.right,T)
else:
T.count = T.count + 1
return T
def inorder(T):
if T is None:
return
else:
inorder(T.left)
b = back(T)
if b:
print("back:",b.data)
print(T.data)
n = next(T)
if n:
print("next:",n.data)
inorder(T.right)
def preorder(T,i=0):
if T is None:
return
else:
for j in range(i):
sys.stdout.write(" ")
print(T.data)
preorder(T.left,i+1)
preorder(T.right,i+1)
def next(node):
if node is None:
return
if node.right:
n = node.right
while n.left:
n = n.left
return n
else:
n = node
while n.parent and n.parent.right is n:
n = n.parent
if n.parent and n.parent.left is n:
return n.parent
else:
return
def back(node):
if node is None:
return
if node.left:
n = node.left
while n.right:
n = n.right
return n
else:
n = node
while n.parent and n.parent.left is n:
n = n.parent
if n.parent and n.parent.right is n:
return n.parent
else:
return
T = None
T = insert(7,T)
T = insert(4,T)
T = insert(2,T)
T = insert(1,T)
T = insert(13,T)
T = insert(15,T)
T = insert(16,T)
T = insert(6,T)
T = insert(5,T)
T = insert(3,T)
T = insert(11,T)
T = insert(14,T)
T = insert(12,T)
T = insert(9,T)
T = insert(8,T)
T = insert(10,T)
T = delete(11,T)
T = delete(12,T)
T = delete(13,T)
T = delete(8,T)
preorder(T)
inorder(T)
输出
7
4
2
1
3
6
5
14
9
10
15
16
1
('next:', 2)
('back:', 1)
2
('next:', 3)
('back:', 2)
3
('next:', 4)
('back:', 3)
4
('next:', 5)
('back:', 4)
5
('next:', 6)
('back:', 5)
6
('next:', 7)
('back:', 6)
7
('next:', 9)
9
('next:', 10)
('back:', 9)
10
('next:', 12)
('back:', 10)
14
('next:', 15)
('back:', 14)
15
('next:', 16)
('back:', 15)
16
预期 - 9 的后面是 7
我的答案
def delete(x,T,parent=None):
if T is None:
print('Element Not Found')
elif x<T.data:
T.left = delete(x,T.left,T)
elif x>T.data:
T.right = delete(x,T.right,T)
elif T.count==1:
# 2 CHILDREN
if T.left and T.right:
TempNode = findMin(T.right)
T.data = TempNode.data
T.right = delete(TempNode.data,T.right,T)
# 0 CHILDREN
elif T.left is None and T.right is None:
T = None
# 1 CHILDREN
elif T.right is not None:
T = T.right
T.parent = parent
elif T.left is not None:
T = T.left
T.parent = parent
else:
T.count = T.count - 1
return T
现在
9
4
2
1
3
5
14
10
15
16
1 (2)
('next:', 2)
('back:', 1)
2 (4)
('next:', 3)
('back:', 2)
3 (2)
('next:', 4)
('back:', 3)
4 (9)
('next:', 5)
('back:', 4)
5 (4)
('next:', 9)
('back:', 5)
9
('next:', 10)
('back:', 9)
10 (14)
('next:', 14)
('back:', 10)
14 (9)
('next:', 15)
('back:', 14)
15 (14)
('next:', 16)
('back:', 15)
16 (15)