[Leetcode]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?

树的后序遍历~同样，先是递归解法~代码如下~

class Solution:    # @param root, a tree node    # @return a list of integers    def postorderTraversal(self, root):        if root is None: return []        self.res = []        self.helper(root)        return self.res            def helper(self, root):        if root is None: return        self.helper(root.left)        self.helper(root.right)        self.res.append(root.val)

非递归解法如下~

1）如果当前栈顶元素的右结点存在并且还没访问过（也就是右结点不等于上一个访问结点），那么就把当前结点移到右结点继续循环；
2）如果栈顶元素右结点是空或者已经访问过，那么说明栈顶元素的左右子树都访问完毕，应该访问自己继续回溯了。

class Solution:    # @param root, a tree node    # @return a list of integers    def postorderTraversal(self, root):        if root is None: return []        res, stack, curr, prev = [], [], root, None        while stack or curr:            if curr:                stack.append(curr)                curr = curr.left            else:                parent = stack[-1]                if parent.right in (None, prev):                    prev = stack.pop()                    res.append(prev.val)                else:                    curr = parent.right        return res

还有Morris遍历法，可以参考这篇博文 http://blog.csdn.net/linhuanmars/article/details/22009351