[英]Why is inorder tree traversal without recursion in Python running infinitely?
我正在尝试在不使用递归的情况下对二叉树进行中序树遍历,但似乎 while 循环一直在无限运行。 任何帮助,将不胜感激。
class Node:
def __init__(self, data):
self.left = None
self.right = None
self.data = data
def inOrder(root):
s = []
while s is not None or root is not None:
if root is not None:
s.append(root.left)
if root.left:
root = root.left
else:
root = s.pop()
print(root.data)
if root.right:
root = root.right
if __name__=='__main__':
root = Node(5)
root.left = Node(3)
root.left.right = Node(2)
root.left.left = Node(4)
root.right = Node(10)
root.right.left = Node(9)
root.right.right = Node(20)
# 5
# / \
# 3 10
# / \ / \
# 4 2 9 20
inOrder(root)
检查以下代码以进行有序遍历:
class Node:
def __init__(self, data):
self.left = None
self.right = None
self.data = data
def inOrder(root):
s = []
s.append(root)
while len(s) > 0: # Check if stack is not empty
if root.left: #Case 1: Traverse left if there is an element left of the current root
s.append(root.left)
root = root.left
else:
root = s.pop() #Case 2: If there is no element on the left, print the current root
print(root.data)
if root.right: #Case 3: If there is an element on the right, traverse right of the current root
s.append(root.right)
root = root.right
if __name__=='__main__':
root = Node(5)
root.left = Node(3)
root.left.right = Node(2)
root.left.left = Node(4)
root.right = Node(10)
root.right.left = Node(9)
root.right.right = Node(20)
inOrder(root)
你的想法是正确的,但问题是
s is not None
不一样
s!=[]
您的 s 是一个始终存在的堆栈,您实际上想检查您的堆栈是否为空。
您总是将s
初始化为一个空列表,该列表永远不会为None
。 您想检查是否为not s
,而not s
是否s is not None
。
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