如何在Python中实现二进制搜索树？
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这是我到目前为止所得到的但它不起作用：
class Node:
rChild,lChild,data = None,None,None
def __init__(self,key):
self.rChild = None
self.lChild = None
self.data = key
class Tree:
root,size = None,0
def __init__(self):
self.root = None
self.size = 0
def insert(self,node,someNumber):
if node is None:
node = Node(someNumber)
else:
if node.data > someNumber:
self.insert(node.rchild,someNumber)
else:
self.insert(node.rchild, someNumber)
return
def main():
t = Tree()
t.root = Node(4)
t.root.rchild = Node(5)
print t.root.data #this works
print t.root.rchild.data #this works too
t = Tree()
t.insert(t.root,4)
t.insert(t.root,5)
print t.root.data #this fails
print t.root.rchild.data #this fails too
if __name__ == '__main__':
main()
答案
以下是二进制插入的快速示例：
class Node:
def __init__(self, val):
self.l_child = None
self.r_child = None
self.data = val
def binary_insert(root, node):
if root is None:
root = node
else:
if root.data > node.data:
if root.l_child is None:
root.l_child = node
else:
binary_insert(root.l_child, node)
else:
if root.r_child is None:
root.r_child = node
else:
binary_insert(root.r_child, node)
def in_order_print(root):
if not root:
return
in_order_print(root.l_child)
print root.data
in_order_print(root.r_child)
def pre_order_print(root):
if not root:
return
print root.data
pre_order_print(root.l_child)
pre_order_print(root.r_child)
r = Node(3)
binary_insert(r, Node(7))
binary_insert(r, Node(1))
binary_insert(r, Node(5))
3
/
1 7
/
5
print "in order:"
in_order_print(r)
print "pre order"
pre_order_print(r)
in order:
1
3
5
7
pre order
3
1
7
5
另一答案
它易于使用两个类实现BST，1。节点和2.树树类仅用于用户界面，实际方法将在Node类中实现。
class Node():
def __init__(self,val):
self.value = val
self.left = None
self.right = None
def _insert(self,data):
if data == self.value:
return False
elif data < self.value:
if self.left:
return self.left._insert(data)
else:
self.left = Node(data)
return True
else:
if self.right:
return self.right._insert(data)
else:
self.right = Node(data)
return True
def _inorder(self):
if self:
if self.left:
self.left._inorder()
print(self.value)
if self.right:
self.right._inorder()
class Tree():
def __init__(self):
self.root = None
def insert(self,data):
if self.root:
return self.root._insert(data)
else:
self.root = Node(data)
return True
def inorder(self):
if self.root is not None:
return self.root._inorder()
else:
return False
if __name__=="__main__":
a = Tree()
a.insert(16)
a.insert(8)
a.insert(24)
a.insert(6)
a.insert(12)
a.insert(19)
a.insert(29)
a.inorder()
用于检查BST是否正确实现的顺序功能。
另一答案
下面的代码是关于@DDing的答案的基础，以及我从类中学到的东西，它使用while循环来插入（在代码中指出）。
class Node:
def __init__(self, val):
self.l_child = None
self.r_child = None
self.data = val
def binary_insert(root, node):
y = None
x = root
z = node
#while loop here
while x is not None:
y = x
if z.data < x.data:
x = x.l_child
else:
x = x.r_child
z.parent = y
if y == None:
root = z
elif z.data < y.data:
y.l_child = z
else:
y.r_child = z
def in_order_print(root):
if not root:
return
in_order_print(root.l_child)
print(root.data)
in_order_print(root.r_child)
r = Node(3)
binary_insert(r, Node(7))
binary_insert(r, Node(1))
binary_insert(r, Node(5))
in_order_print(r)
另一答案
这是一个有效的解决方案。
class BST:
def __init__(self,data):
self.root = data
self.left = None
self.right = None
def insert(self,data):
if self.root == None:
self.root = BST(data)
elif data > self.root:
if self.right == None:
self.right = BST(data)
else:
self.right.insert(data)
elif data < self.root:
if self.left == None:
self.left = BST(data)
else:
self.left.insert(data)
def inordertraversal(self):
if self.left != None:
self.left.inordertraversal()
print (self.root),
if self.right != None:
self.right.inordertraversal()
t = BST(4)
t.insert(1)
t.insert(7)
t.insert(3)
t.insert(6)
t.insert(2)
t.insert(5)
t.inordertraversal()
另一答案
接受的答案忽略了为插入的每个节点设置父属性，否则无法实现successor方法，该方法在O（h）时间内在有序树遍历中找到后继，其中h是树的高度（如反对步行所需的O（n）时间。
这是一个基于Cormen等人，算法导论中给出的伪代码的实现，包括分配parent属性和successor方法：
class Node(object):
def __init__(self, key):
self.key = key
self.left = None
self.right = None
self.parent = None
class Tree(object):
def __init__(self, root=None):
self.root = root
def insert(self, z):
y = None
x = self.root
while x is not None:
y = x
if z.key < x.key:
x = x.left
else:
x = x.right
z.parent = y
if y is None:
self.root = z # Tree was empty
elif z.key < y.key:
y.left = z
else:
y.right = z
@staticmethod
def minimum(x):
while x.left is not None:
x = x.left
return x
@staticmethod
def successor(x):
if x.right is not None:
return Tree.minimum(x.right)
y = x.parent
while y is not None and x == y.right:
x = y
y = y.parent
return y
以下是一些测试，表明树的行为符合DTing给出的示例：
import pytest
@pytest.fixture
def tree():
t = Tree()
t.insert(Node(3))
t.insert(Node(1))
t.insert(Node(7))
t.insert(Node(5))
return t
def test_tree_insert(tree):
assert tree.root.key == 3
assert tree.root.left.key == 1
assert tree.root.right.key == 7
assert tree.root.right.left.key == 5
def test_tree_successor(tree):
assert Tree.successor(tree.root.left).key == 3
assert Tree.successor(tree.root.right.left).key == 7
if __name__ == "__main__":
pytest.main([__file__])
另一答案
问题或代码中至少有一个问题在于： -
def insert(self,node,someNumber):
if node is None:
node = Node(someNumber)
else:
if node.data > someNumber:
self.insert(node.rchild,someNumber)
else:
self.insert(node.rchild, someNumber)
return
您会看到语句“if node.data> someNumber：”和关联的“else：”语句后面都有相同的代码。即if语句是真还是假，你做同样的事情。
我建议你可能打算在这里做不同的事情，也许其中一个应该说self.insert（node.lchild，someNumber）？
另一答案
另一个Python BST解决方案
class Node(object):
def __init__(self, value):
self.left_node = None
self.right_node = None
self.value = value
def __str__(self):
return "[%s, %s, %s]" % (self.left_node, self.value, self.right_node)
def insertValue(self, new_value):
"""
1. if current Node doesnt have value then assign to self
2. new_value lower than current Node's value then go left
2. new_value greater than current Node's value then go right
:return:
"""
if self.value:
if new_value < self.value:
# add to left
if self.left_node is None: # reached start add value to start
self.left_node = Node(new_value)
else:
self.left_node.insertValue(new_value) # search
elif new_value > self.value:
# add to right
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