为什么在Python中无限运行无递归的中序树遍历？

Question

I'm trying to do Inorder tree traversal for binary trees without using recursion but it seems like the while loop keeps running infinitely. Any help would be appreciated.

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)

Answer 1

Check the following code for inorder traversal:

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)

Answer 2

Your idea is correct but the problem is that

s is not None

Is not the same as

s!=[]

Your s is a stack which is always there and you actually want to check whether your stack is empty or not.

Answer 3

You are always initiating s to an empty list, which will never be None . You wanted to check whether not s , not whether s is not None .