【LEETCODE】145-Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
   1
    \
     2
    /
   3
return [3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
참고:
http://bookshadow.com/weblog/2015/01/19/leetcode-binary-tree-postorder-traversal/

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        
        if root is None:
            return []
        
        ans = []                        ＃    
        stack = [root]                  ＃stack  cur   
        pre = None                      ＃pre  cur   
        
        while stack:
            
            cur = stack[-1]
            
            if pre is None or pre.left==cur or pre.right==cur:       ＃ pre cur parent ，cur    child ，   child   child
                if cur.left:
                    stack.append(cur.left)
                elif cur.right:
                    stack.append(cur.right)
            elif pre==cur.left and cur.right:                        ＃ pre cur  child ，  cur  child ，cur     
                stack.append(cur.right)
            else:                                       ＃ pre＝＝cur ，  pre＝＝cur.left  cur.right    ，
                ans.append(cur.val)                     ＃ans  cur.val
                stack.pop()                             ＃stack     
            
            pre=cur           <span style="font-family: Arial, Helvetica, sans-serif;">                     ＃pre  cur   </span>

        
        return ans