如何遍历python中的二叉搜索树（无递归）

Question

So far I have

def tree_iterate():
parent, current = None, self.root
lst = []
while current is not None: 
 if current.left not None:
  lst.append(current.item) 
  parent, current = current, current.left
 if current.right not None:
  lst.append(current.item)
  parent, current = current, current.right

(sorry about spacing I'm quite new at this)

I'm not quite sure how to iterate on both sides of the tree when current has left and right, without using recursion. My main goal is to have a list of all the nodes in this BST enter code here

Answer 1

To get a list of all nodes in the BST iteratively, use Breadth-First Search (BFS). Note that this won't give you the nodes in sorted order:

queue = [root]
result = []
while queue:
    l = queue.pop(0)
    result.append(l)

    if l.left != None:
        queue.append(l.left)
    if l.right!= None:
        queue.append(l.right)

If you want the nodes in sorted order, you will need to simulate inorder traversal using a stack:

result = []
stack = [root]
while stack:
    stack[-1].visited = True
    if stack[-1].left != None and not stack[-1].left.visited:
        stack.append(stack[-1].left)
    else:
        node = stack.pop()
        result.append(node)
        if stack[-1].right != None:
            stack.append(stack[-1].right)