python二叉树练习

时间:2023-03-09 22:22:32
python二叉树练习
#coding=utf8
node_list=[5,3,6,2,4,None,8,1,None,None,None,7,9] class Node:
def __init__(self,item):
self.item = item
self.child1 = None
self.child2 = None class Tree:
def __init__(self):
self.root = None def add(self, item):
node = Node(item)
if self.root is None:
self.root = node
else:
q = [self.root] while True:
pop_node = q.pop(0)
if pop_node.child1 is None:
pop_node.child1 = node
return
elif pop_node.child2 is None:
pop_node.child2 = node
return
else:
q.append(pop_node.child1)
q.append(pop_node.child2) def traverse(self): # 层次遍历
if self.root is None:
return None
q = [self.root]
res = [self.root.item]
while q != []:
pop_node = q.pop(0)
if pop_node.child1 is not None:
q.append(pop_node.child1)
res.append(pop_node.child1.item) if pop_node.child2 is not None:
q.append(pop_node.child2)
res.append(pop_node.child2.item)
return res def preorder(self,root): # 先序遍历
if root is None:
return []
result = [root.item]
left_item = self.preorder(root.child1)
right_item = self.preorder(root.child2)
return result + left_item + right_item def inorder(self,root): # 中序序遍历
if root is None:
return []
result = [root.item]
left_item = self.inorder(root.child1)
right_item = self.inorder(root.child2)
return left_item + result + right_item def postorder(self,root): # 后序遍历
if root is None:
return []
result = [root.item]
left_item = self.postorder(root.child1)
right_item = self.postorder(root.child2)
return left_item + right_item + result t = Tree()
for i in node_list:
t.add(i)
print('层序遍历:',t.traverse())
#print('先序遍历:',t.preorder(t.root))
#print('中序遍历:',t.inorder(t.root))
print('后序遍历:',t.postorder(t.root))