leetcode_145. 二叉树的后序遍历

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