二叉树的构造和遍历——python实现

question

使用0,1,2,3,4,5,6,7,8,9十个数构造一个二叉树(非排序)。并进行层次遍历、前序遍历、中序遍历和后序遍历。

solution

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class :
def __init__(self,item):
self.item = item
self.child1 = None
self.child2 = None
class Tree: # 树类
def __init__(self):
self.root = None
def add(self, item): # 构造二叉树
node = Node(item)
if self.root is None:
self.root = node
else:
q = [self.root]
while q:
pop_node = q.pop(0)
if pop_node.child1 is None:
pop_node.child1 = node
return # return跳出循环
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 range(10):
t.add(i)
print('层次遍历:',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))