[英]Printing out BST nodes in order - only printing first node?
I expect to print out [6,5,10
我希望打印出
[6,5,10
instead I only get 10
: 相反,我只得到
10
:
class BST:
arr = []
def __init__(self):
self.height = 0
self.key = 0
self.left = None
self.right = None
def __str__(self):
return str(self.key)
def populate(self):
print("populating")
print(self.key)
if (self.left != None):
BST.arr = BST.arr + [self.left.populate()]
if (self.right != None):
BST.arr = BST.arr + [self.right.populate()]
return self.key
m1 = BST()
m1.key = 10
m2 = BST()
m2.key = 5
m1.left = m2
print(m1.left != None)
m3 = BST()
m3.key = 6
m2.left = m3
res = m1.populate()
print(res)
Here's a function that does in order traversal: 这是一个按顺序遍历的函数:
def inOrder(node, order):
if node is not None:
inOrder(node.left, order)
order.append(node.key)
inOrder(node.right, order)
You can write a wrapper function that does return a list and use it like this: 您可以编写一个包装函数,该函数确实返回一个列表并像这样使用它:
def inOrder(root):
order = []
inOrder(root, order)
return order
Edit: Or you can do that with one function like this: 编辑:或者您可以使用以下一种功能来做到这一点:
def inOrder(node):
if node is None:
return []
return inOrder(node.left) + [node.key] + inOrder(node.right)
A workaround is to update your populate
method: 解决方法是更新您的
populate
方法:
class BST:
arr = []
def __init__(self):
self.height = 0
self.key = 0
self.left = None
self.right = None
def __str__(self):
return str(self.key)
def populate(self):
self.arr = []
temp = self
while temp.left != None or temp.right!=None:
self.arr.append(temp.key)
if temp.left!=None:
temp = temp.left
elif temp.right!=None:
temp = temp.right
if temp.left == None and temp.right==None:
self.arr.append(temp.key)
return self.arr
m1 = BST()
m1.key = 10
m2 = BST()
m2.key = 5
m1.left = m2
print(m1.left != None)
m3 = BST()
m3.key = 6
m2.left = m3
res = m1.populate()
print(res)
Output: 输出:
True
[10, 5, 6]
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