题解 | #二叉树的后序遍历#

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#
# 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
#
# 
# @param root TreeNode类 
# @return int整型一维数组
#
class Solution:
    def postorder(self, list: List[int], root: TreeNode):
        # 遇到空节点则返回
        if not root:
            return
        # 先去左子树
        self.postorder(list, root.left) 
        # 再去右子树
        self.postorder(list, root.right) 
        # 最后访问根节点
        list.append(root.val) 
        
    def postorderTraversal(self , root: TreeNode) -> List[int]:
        res = []
        # 递归后序遍历
        self.postorder(res, root) 
        return res

递归问题还是不太明白,前序遍历\中序遍历和后续遍历对比回顾.