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 0 :(得分：54)
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
答案 1 :(得分：10)
class Node:
rChild,lChild,data = None,None,None
这是错误的 - 它使您的变量类变量 - 也就是说，每个Node实例都使用相同的值（更改任何节点的rChild会为所有节点更改它！）。这显然不是你想要的;尝试
class Node:
def __init__(self, key):
self.rChild = None
self.lChild = None
self.data = key
现在每个节点都有自己的一组变量。这同样适用于您对Tree的定义，
class Tree:
root,size = None,0 # <- lose this line!
def __init__(self):
self.root = None
self.size = 0
此外，每个类应该是一个派生自“object”类的“new-style”类，并且应该链回到object .__ init __（）：
class Node(object):
def __init__(self, data, rChild=None, lChild=None):
super(Node,self).__init__()
self.data = data
self.rChild = rChild
self.lChild = lChild
class Tree(object):
def __init__(self):
super(Tree,self).__init__()
self.root = None
self.size = 0
另外，main（）缩进太多 - 如图所示，它是Tree的一种方法，因为它不接受 self 参数而不可调用。
此外，您正在直接修改对象的数据（t.root = Node(4)）哪种类型的破坏封装（首先是拥有类的全部内容）;你应该做更像
def main():
t = Tree()
t.add(4) # <- let the tree create a data Node and insert it
t.add(5)
答案 2 :(得分：7)
class BST:
def __init__(self, val=None):
self.left = None
self.right = None
self.val = val
def __str__(self):
return "[%s, %s, %s]" % (self.left, str(self.val), self.right)
def isEmpty(self):
return self.left == self.right == self.val == None
def insert(self, val):
if self.isEmpty():
self.val = val
elif val < self.val:
if self.left is None:
self.left = BST(val)
else:
self.left.insert(val)
else:
if self.right is None:
self.right = BST(val)
else:
self.right.insert(val)
a = BST(1)
a.insert(2)
a.insert(3)
a.insert(0)
print a
答案 3 :(得分：5)
class Node:
rChild,lChild,parent,data = None,None,None,0
def __init__(self,key):
self.rChild = None
self.lChild = None
self.parent = None
self.data = key
class Tree:
root,size = None,0
def __init__(self):
self.root = None
self.size = 0
def insert(self,someNumber):
self.size = self.size+1
if self.root is None:
self.root = Node(someNumber)
else:
self.insertWithNode(self.root, someNumber)
def insertWithNode(self,node,someNumber):
if node.lChild is None and node.rChild is None:#external node
if someNumber > node.data:
newNode = Node(someNumber)
node.rChild = newNode
newNode.parent = node
else:
newNode = Node(someNumber)
node.lChild = newNode
newNode.parent = node
else: #not external
if someNumber > node.data:
if node.rChild is not None:
self.insertWithNode(node.rChild, someNumber)
else: #if empty node
newNode = Node(someNumber)
node.rChild = newNode
newNode.parent = node
else:
if node.lChild is not None:
self.insertWithNode(node.lChild, someNumber)
else:
newNode = Node(someNumber)
node.lChild = newNode
newNode.parent = node
def printTree(self,someNode):
if someNode is None:
pass
else:
self.printTree(someNode.lChild)
print someNode.data
self.printTree(someNode.rChild)
def main():
t = Tree()
t.insert(5)
t.insert(3)
t.insert(7)
t.insert(4)
t.insert(2)
t.insert(1)
t.insert(6)
t.printTree(t.root)
if __name__ == '__main__':
main()
我的解决方案。
答案 4 :(得分：5)
Op的Tree.insert方法有资格获得“本周大赢家”奖 - 它不会插入任何内容。它创建一个没有连接到任何其他节点的节点（不是有任何节点将它连接到），然后在方法返回时删除创建的节点。
为了启发@Hugh Bothwell：
>>> class Foo(object):
... bar = None
...
>>> a = Foo()
>>> b = Foo()
>>> a.bar
>>> a.bar = 42
>>> b.bar
>>> b.bar = 666
>>> a.bar
42
>>> b.bar
666
>>>
答案 5 :(得分：2)
我发现insert部分的解决方案有点笨拙。您可以返回root引用并稍微简化一下：
def binary_insert(root, node):
if root is None:
return node
if root.data > node.data:
root.l_child = binary_insert(root.l_child, node)
else:
root.r_child = binary_insert(root.r_child, node)
return root
答案 6 :(得分：1)
只是帮助你开始的东西。
（简单的想法）二叉树搜索很可能是在python中实现的：
def search(node, key):
if node is None: return None # key not found
if key< node.key: return search(node.left, key)
elif key> node.key: return search(node.right, key)
else: return node.value # found key
现在你只需要实现脚手架（树创建和值插入），你就完成了。
答案 7 :(得分：1)
另一个带有排序键的Python BST（默认为值）
LEFT = 0
RIGHT = 1
VALUE = 2
SORT_KEY = -1
class BinarySearchTree(object):
def __init__(self, sort_key=None):
self._root = []
self._sort_key = sort_key
self._len = 0
def insert(self, val):
if self._sort_key is None:
sort_key = val // if no sort key, sort key is value
else:
sort_key = self._sort_key(val)
node = self._root
while node:
if sort_key < node[_SORT_KEY]:
node = node[LEFT]
else:
node = node[RIGHT]
if sort_key is val:
node[:] = [[], [], val]
else:
node[:] = [[], [], val, sort_key]
self._len += 1
def minimum(self):
return self._extreme_node(LEFT)[VALUE]
def maximum(self):
return self._extreme_node(RIGHT)[VALUE]
def find(self, sort_key):
return self._find(sort_key)[VALUE]
def _extreme_node(self, side):
if not self._root:
raise IndexError('Empty')
node = self._root
while node[side]:
node = node[side]
return node
def _find(self, sort_key):
node = self._root
while node:
node_key = node[SORT_KEY]
if sort_key < node_key:
node = node[LEFT]
elif sort_key > node_key:
node = node[RIGHT]
else:
return node
raise KeyError("%r not found" % sort_key)
答案 8 :(得分：1)
这是一个紧凑的，面向对象的递归实现：
class BTreeNode(object):
def __init__(self, data):
self.data = data
self.rChild = None
self.lChild = None
def __str__(self):
return (self.lChild.__str__() + '<-' if self.lChild != None else '') + self.data.__str__() + ('->' + self.rChild.__str__() if self.rChild != None else '')
def insert(self, btreeNode):
if self.data > btreeNode.data: #insert left
if self.lChild == None:
self.lChild = btreeNode
else:
self.lChild.insert(btreeNode)
else: #insert right
if self.rChild == None:
self.rChild = btreeNode
else:
self.rChild.insert(btreeNode)
def main():
btreeRoot = BTreeNode(5)
print 'inserted %s:' %5, btreeRoot
btreeRoot.insert(BTreeNode(7))
print 'inserted %s:' %7, btreeRoot
btreeRoot.insert(BTreeNode(3))
print 'inserted %s:' %3, btreeRoot
btreeRoot.insert(BTreeNode(1))
print 'inserted %s:' %1, btreeRoot
btreeRoot.insert(BTreeNode(2))
print 'inserted %s:' %2, btreeRoot
btreeRoot.insert(BTreeNode(4))
print 'inserted %s:' %4, btreeRoot
btreeRoot.insert(BTreeNode(6))
print 'inserted %s:' %6, btreeRoot
上述main（）的输出是：
inserted 5: 5
inserted 7: 5->7
inserted 3: 3<-5->7
inserted 1: 1<-3<-5->7
inserted 2: 1->2<-3<-5->7
inserted 4: 1->2<-3->4<-5->7
inserted 6: 1->2<-3->4<-5->6<-7
答案 9 :(得分：1)
使用两个类很容易实现BST，1。节点和2.树 Tree类将仅用于用户界面，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是否正确实施的Inorder函数。
答案 10 :(得分：0)
一种简单的递归方法，仅具有1个函数并使用值数组：
class TreeNode(object):
def __init__(self, value: int, left=None, right=None):
super().__init__()
self.value = value
self.left = left
self.right = right
def __str__(self):
return str(self.value)
def create_node(values, lower, upper) -> TreeNode:
if lower > upper:
return None
index = (lower + upper) // 2
value = values[index]
node = TreeNode(value=value)
node.left = create_node(values, lower, index - 1)
node.right = create_node(values, index + 1, upper)
return node
def print_bst(node: TreeNode):
if node:
# Simple pre-order traversal when printing the tree
print("node: {}".format(node))
print_bst(node.left)
print_bst(node.right)
if __name__ == '__main__':
vals = [0, 1, 2, 3, 4, 5, 6]
bst = create_node(vals, lower=0, upper=len(vals) - 1)
print_bst(bst)
如您所见，我们实际上只需要一种方法，该方法是递归的：create_node。我们在每个values方法调用中传递完整的create_node数组，但是，每次进行递归调用时，我们都会更新lower和upper索引值。
然后，使用lower和upper索引值，计算当前节点的index值并将其捕获在value中。 value是当前节点的值，我们用它来创建节点。
从那里，我们通过递归调用函数来设置left和right的值，直到lower大于{{1 }}。
重要提示：我们在创建树的upper边时更新upper的值。相反，我们在创建树的left边时更新lower的值。
希望这会有所帮助！
答案 11 :(得分：0)
class TreeNode:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
class BinaryTree:
def __init__(self, root=None):
self.root = root
def add_node(self, node, value):
"""
Node points to the left of value if node > value; right otherwise,
BST cannot have duplicate values
"""
if node is not None:
if value < node.value:
if node.left is None:
node.left = TreeNode(value)
else:
self.add_node(node.left, value)
else:
if node.right is None:
node.right = TreeNode(value)
else:
self.add_node(node.right, value)
else:
self.root = TreeNode(value)
def search(self, value):
"""
Value will be to the left of node if node > value; right otherwise.
"""
node = self.root
while node is not None:
if node.value == value:
return True # node.value
if node.value > value:
node = node.left
else:
node = node.right
return False
def traverse_inorder(self, node):
"""
Traverse the left subtree of a node as much as possible, then traverse
the right subtree, followed by the parent/root node.
"""
if node is not None:
self.traverse_inorder(node.left)
print(node.value)
self.traverse_inorder(node.right)
def main():
binary_tree = BinaryTree()
binary_tree.add_node(binary_tree.root, 200)
binary_tree.add_node(binary_tree.root, 300)
binary_tree.add_node(binary_tree.root, 100)
binary_tree.add_node(binary_tree.root, 30)
binary_tree.traverse_inorder(binary_tree.root)
print(binary_tree.search(200))
if __name__ == '__main__':
main()
答案 12 :(得分：0)
def BinaryST(list1,key):
start = 0
end = len(list1)
print("Length of List: ",end)
for i in range(end):
for j in range(0, end-i-1):
if(list1[j] > list1[j+1]):
temp = list1[j]
list1[j] = list1[j+1]
list1[j+1] = temp
print("Order List: ",list1)
mid = int((start+end)/2)
print("Mid Index: ",mid)
if(key == list1[mid]):
print(key," is on ",mid," Index")
elif(key > list1[mid]):
for rindex in range(mid+1,end):
if(key == list1[rindex]):
print(key," is on ",rindex," Index")
break
elif(rindex == end-1):
print("Given key: ",key," is not in List")
break
else:
continue
elif(key < list1[mid]):
for lindex in range(0,mid):
if(key == list1[lindex]):
print(key," is on ",lindex," Index")
break
elif(lindex == mid-1):
print("Given key: ",key," is not in List")
break
else:
continue
size = int(input("Enter Size of List: "))
list1 = []
for e in range(size):
ele = int(input("Enter Element in List: "))
list1.append(ele)
key = int(input("\nEnter Key for Search: "))
print("\nUnorder List: ",list1)
BinaryST(list1,key)
答案 13 :(得分：0)
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
if self.right_node is None: # reached end add value to end
self.right_node = Node(new_value)
else:
self.right_node.insertValue(new_value) # search
else:
self.value = new_value
def findValue(self, value_to_find):
"""
1. value_to_find is equal to current Node's value then found
2. if value_to_find is lower than Node's value then go to left
3. if value_to_find is greater than Node's value then go to right
"""
if value_to_find == self.value:
return "Found"
elif value_to_find < self.value and self.left_node:
return self.left_node.findValue(value_to_find)
elif value_to_find > self.value and self.right_node:
return self.right_node.findValue(value_to_find)
return "Not Found"
def printTree(self):
"""
Nodes will be in sequence
1. Print LHS items
2. Print value of node
3. Print RHS items
"""
if self.left_node:
self.left_node.printTree()
print(self.value),
if self.right_node:
self.right_node.printTree()
def isEmpty(self):
return self.left_node == self.right_node == self.value == None
def main():
root_node = Node(12)
root_node.insertValue(6)
root_node.insertValue(3)
root_node.insertValue(7)
# should return 3 6 7 12
root_node.printTree()
# should return found
root_node.findValue(7)
# should return found
root_node.findValue(3)
# should return Not found
root_node.findValue(24)
if __name__ == '__main__':
main()
答案 14 :(得分：0)
问题或代码中至少有一个问题在这里：-
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“ if node.data> someNumber：”语句和关联的“ else：”语句后面都有相同的代码。也就是说，无论if语句为true还是false，您都会做同样的事情。
我建议您可能打算在这里做不同的事情，也许其中之一应该说self.insert（node.lchild，someNumber）吗？
答案 15 :(得分：0)
接受的答案忽略为插入的每个节点设置父属性，否则无法实现{<1}}方法，该方法在 O 中的有序树遍历中找到后继方法（ h ）时间，其中<em> h 是树的高度（与 O （ n ）时间相对步行所需。）
这是基于Cormen等人提供的伪代码算法简介的实现，包括successor属性和parent方法的分配：
successor以下是一些测试，表明树的行为符合DTing给出的示例的预期：
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
答案 16 :(得分：0)
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()
答案 17 :(得分：0)
以下代码是基于@DDing的答案和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)
答案 18 :(得分：-1)
here是一个示例实现。