python 二叉树遍历 非递归遍历

class Node(object):
def init(self,val=None,left=None,right=None):
# val为节点的值，left为左子树的根节点，right为右子树的根节点
self.val = val
self.left = left
self.right = right
class BT(object):
def init(self):
self.item = []
#先序遍历
def preorder(self,root):
stack = []
while root or stack:
while root:
self.item.append(root.val)
stack.append(root)
root = root.left
if stack:
item = stack.pop()
root = item.right
#中序遍历
def midorder(self,root):
stack = []
while root or stack:
while root:
stack.append(root)
root = root.left
if stack:
item = stack.pop()
self.item.append(item.val)
root = item.right
#后序遍历，思想就是用stack1去按照根、右、左的顺序去遍历，stack2就是按照这个顺序去打印，stack2打印的顺序刚好时右序遍历的倒序
def postorder(self,root):
stack1 = []
# 后序的倒序，后 右 左
stack2 = []
while root or stack1:
while root:
stack2.append(root.val)
stack1.append(root)
root = root.right
item = stack1.pop()
root = item.left
self.item = [stack2[-i] for i in range(1,len(stack2)+1)]

bt = Node(‘D’,left=Node(‘A’,left=Node(‘F’),right=Node(‘G’,right=Node(‘E’))),right=Node(‘C’,left=Node(‘K’),right=Node(‘M’,left=Node(‘S’))))
tree1 = BT()
tree1.preorder(bt)
print(tree1.item)
tree1.midorder(bt)
print(tree1.item)
tree1.postorder(bt)
print(tree1.item)