题解 | #二叉树的后序遍历# #非DFS，挺难的

#coding:utf-8
# 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 __init__(self):
        self.path = []
        self.stack = []

    def postorderTraversal(self , root ):
        # write code here
        #dfs
        #self.dfs(root)
        #travel 
        self.travel(root)
        print ("Path: ", self.path)
        return self.path

    def travel(self, node):
        if node == None:
            return
        last = node
        self.stack.append(node)
        while len(self.stack) > 0:
            peek = self.stack[-1]
            if peek.left != None and peek.left != last and peek.right != last:
                self.stack.append(peek.left)
            elif peek.right != None and peek.right != last:
                self.stack.append(peek.right)
            else:
                #last = self.stack.pop()
                self.path.append(self.stack.pop().val)
                last = peek
        return


    def dfs(self, node):
        if node == None:
            return
        self.dfs(node.left)
        self.dfs(node.right)
        self.path.append(node.val)
        return