為什么在Python中無限運行無遞歸的中序樹遍歷？

Question

我正在嘗試在不使用遞歸的情況下對二叉樹進行中序樹遍歷，但似乎 while 循環一直在無限運行。 任何幫助，將不勝感激。

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)

Answer 1

檢查以下代碼以進行有序遍歷：

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

你的想法是正確的，但問題是

s is not None

不一樣

s!=[]

您的 s 是一個始終存在的堆棧，您實際上想檢查您的堆棧是否為空。

Answer 3

您總是將s初始化為一個空列表，該列表永遠不會為None 。 您想檢查是否為not s ，而not s是否s is not None 。