递归

# 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: Optional[TreeNode]) -> List[int]:
        # Solution 1: 递归
        res = []
        def dfs(root:Optional[TreeNode]):
            if root == None:
                return
            dfs(root.left)
            dfs(root.right)
            res.append(root.val)
            
        dfs(root)
        return res   

迭代：和中序遍历、前序遍历略有不同

增加一个辅助变量preNode，记录上一轮访问的节点，来判断当前节点的right是不是上一轮被访问的节点

# 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: Optional[TreeNode]) -> List[int]:
       # Solution 2: 迭代
        res = []
        stack = []
        preNode = None
        while root!=None or stack!=[]:
            while root:
                stack.append(root)
                root = root.left
            root = stack.pop()
            # Note(1)
            if root.right == None or root.right == preNode:
                res.append(root.val)
                preNode = root
                root = None
            else:
            	# Note(2)如果当前节点的右子树还没有被访问过，那么当前节点还要再次入栈
                stack.append(root)
                root = root.right
        return res