145. 二叉树的后序遍历

题目描述

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？

我的解法——递归（AC）

# 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 getpre(self, tree):
        if(not tree):
            return []
        left = self.getpre(tree.left)
        right = self.getpre(tree.right)
        return left + right + [tree.val, ]
    
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        return self.getpre(root)

我的解法——双栈非递归（AC）

# 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(not root):
            return []
        
        s1 = [root, ]
        s2 = []
        
        
        while(len(s1)):
            now = s1[-1]
            s2.append(s1[-1].val)
            s1 = s1[:-1]
            if(now.left):
                s1.append(now.left)
            if(now.right):
                s1.append(now.right)
        
        return s2[::-1]

我的解法——单栈非递归

• 见《程序员代码面试指南》P92