python實現二叉樹

初學python，需要實現乙個決策樹，首先實踐一下利用python實現乙個二叉樹資料結構。建樹的時候做了處理，保證建立的二叉樹是平衡二叉樹。

#

-*- coding: utf-8 -*-
from collections import
deque
class
node:
def__init__(self,val,left=none,right=none):
self.val=val
self.left=left
self.right=right
#setter and getter
defget_val(self):
return
self.val
defset_val(self,val):
self.val=val
defget_left(self):
return
self.left
defset_left(self,left):
self.left=left
defget_right(self):
return
self.right
defset_right(self,right):
self.right=right
class
tree:
def__init__
(self,list):
list=sorted(list)
self.root
=self.build_tree(list)
#遞迴建立平衡二叉樹
defbuild_tree(self,list):
l=0r=len(list)-1
if(l>r):
return
none
if(l==r):
return
node(list[l])
mid=(l+r)/2root=node(list[mid])
root.left=self.build_tree(list[:mid])
root.right=self.build_tree(list[mid+1:])
return
root
#前序遍歷
defpreorder(self,root):
if(root is
none):
return
print
root.val
self.preorder(root.left)
self.preorder(root.right)
#後序遍歷
defpostorder(self,root):
if(root is
none):
return
self.postorder(root.left)
self.postorder(root.right)
print
root.val
#中序遍歷
definorder(self,root):
if(root is
none):
return
self.inorder(root.left)
print
root.val
self.inorder(root.right)
#層序遍歷
deflevelorder(self,root):
if root is
none:
return
queue =deque([root])
while(len(queue)>0):
size=len(queue)
for i in
range(size):
node =queue.popleft()
print
node.val
if node.left is
notnone:
if node.right is
notnone:
list=[1,-1,3,4,5]
tree=tree(list)
print
'中序遍歷：
'tree.inorder(tree.root)
print
'層序遍歷：
'tree.levelorder(tree.root)
print
'前序遍歷：
'tree.preorder(tree.root)
print
'後序遍歷：
'tree.postorder(tree.root)

345層序遍歷3-1

415前序遍歷3-1

145後序遍歷1-1

543建立的二叉樹如下圖所示：

python 二叉樹查詢 Python二叉樹搜尋

stack depth is initialised to 0 def find in tree node,find condition,stack depth assert stack depth max stack depth deeper than max depth stack depth ...

python實現平衡二叉樹 平衡二叉樹python

stack depth is initialised to 0 def find in tree node,find condition,stack depth assert stack depth max stack depth deeper than max depth stack depth ...

python實現二叉樹翻轉

題 給定一棵二叉樹，要求輸出其左右翻轉後二叉樹的中序遍歷。例 翻轉前 翻轉後 1 1 2 3 3 2 4 5 5 4思路 映象翻 只需要遍歷二叉樹，每次訪問乙個結點時，交換其左右孩子，然後遞迴的翻轉左右孩子。class node object def init self,val none lchil...