给定一个二叉树，返回它的 后序 遍历。
示例:
输入: [1,null,2,3]
1
\
2
/
3
输出: [3,2,1]
进阶: 递归算法很简单，你可以通过迭代算法完成吗？
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-postorder-traversal
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
if not root:return []
return self.postorderTraversal(root.left)+self.postorderTraversal(root.right)+[root.val]
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
def postOrder(root:TreeNode)->None:
if not root :return
postOrder(root.left)
postOrder(root.right)
ls.append(root.val)
ls=[]
postOrder(root)
return ls
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def postorderTraversal(self, root: TreeNode) -> List[int]:
if not root:
return []
ls=[]
stack=[]
out=[]
cur=root
while stack or cur:
if cur:
stack.append(cur)
out.append(cur)
cur=cur.right
else:
cur=stack.pop()
cur=cur.left
while(out):
t=out.pop()
ls.append(t.val)
return ls
