Force deduction binary tree middle order traversal (non recursive) Python

# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right

The test system has defined the nodes , Node value is int type , without L/R, Its value is None Not the example null

My general idea flow is ：

1. (while) All the way into the left subtree ( At the same time, its parent node is put on the stack ), Until you see the left leaf node and record result until
2. Put the current node's val Record in result
3. (while) All the way out of the stack there is no right child node , Record while taking val
4. When the node has a right child , If it is a leaf node, record it in result, Not just as an access point for the next loop

stack = []
result = []
visit = root
while visit:
while visit.left:
# When there is a left child
if isinstance(visit.left, int):
# Left the child -> leaf
result.append(visit.left)
# Access the left node
else:
# Left the child -> subtree
stack.append(visit)
visit = visit.left
# Stack current node , Visit left child
result.append(visit.val)
# Access root
while stack and not visit.right:
# Stack is not empty. 、 The visited node has no right child
visit = stack.pop()
result.append(visit.val)
if isinstance(visit.right, int):
# The right child -> leaf
result.append(visit.right)
else:
# The right child -> subtree
visit = visit.right
return result

The test of the buckle is not accurate , Many tests fluctuate a little