为什么Python中无递归的有序树遍历会无限运行？

我试图在不使用递归的情况下对二叉树进行有序树遍历，但while循环似乎一直在无限地运行。任何帮助都将不胜感激

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)

---

检查以下代码中的顺序遍历：

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)

---

您总是将

s

初始化为一个空列表，该列表永远不会是

None

。您想检查

是否不是s

，而不是

s是否不是None

不是解决方案，但是

s

将永远不会是

None

。我也没有附加初始根节点，另一个答案指出了这一点。必须选择它作为正确的一个。不过，非常感谢你的帮助。