Binary Tree Postorder Traversal

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题目描述

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?

思路

递归方法很简单不说了。非递归方法利用栈，在一个while循环里将所有左孩子加入ans结果中，然后再加入右子树，直到dfs为空。我使用了一个pre的list来记录dfs中对应节点的父亲节点，然后在每个数据pop出来之后将其设置为null。这样其实改变的原来的树了。有其他方法可以不用改变原来的树的，这里我没有再多考虑了。

代码(Python)

递归方法：

# Definition for a  binary tree node# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    # @param root, a tree node    # @return a list of integers    def postorderTraversal(self,root):        ans = []        if root is None:            return []        if root.left is not None:            ans+=self.postorderTraversal(root.left)        if root.right is not None:            ans+=self.postorderTraversal(root.right)        ans+= [root.val]        return ans

非递归方法：

# Definition for a  binary tree node# class TreeNode:#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution:    # @param root, a tree node    # @return a list of integers    def postorderTraversal(self,root):        ans = []        dfs = []        pre = []        if root is not None:            dfs.append(root)            pre.append(root)            v = root            p = root        while dfs != []:            while v.left is not None:                dfs.append(v.left)                pre.append(v)                v = v.left            if v.right is not None:                dfs.append(v.right)                pre.append(v)                v = v.right            if v.left is None and v.right is None:                v = dfs.pop()                p = pre.pop()                ans.append(v.val)                if p.left is not None:                    p.left = None                elif p.right is not None:                    p.right = None                if dfs != []:                    v = dfs.pop()                    dfs.append(v)        return ans

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