参考：http://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/
<pre name="code" class="plain">1. Initialize current as root
2. While current is not NULL
If current does not have left child
a) Print current’s data
b) Go to the right child, i.e., current = current->right
Else
a) Make current as right child of the rightmost node in current's left subtree
b) Go to this left child, i.e., current = current->left
When we do 2.Else, for the first round it will execute 2.Else.a/b. When finish the steps and return back,we check again and reset the rightmost node in current's left subtree as null. That is the following statement:
a') Find the rightmost node in current's left subtree (its right child is current) and set its right child as null
b') Go to the right child
cur = root
while cur is not None:
if cur.left is None:
cur = cur.right
else:
ptr = cur.left
while ptr.right != None and ptr.right != cur:
ptr = ptr.right
if ptr.right == None: //Else a/b, the rightmost node in current's left subtree
ptr.right = cur
cur = cur.left
else: //Else a'/b', restore the right child of the right most node in current's left subtree
ptr.right = None
cur = cur.right题目举例：
https://oj.leetcode.com/problems/recover-binary-search-tree/
https://oj.leetcode.com/problems/validate-binary-search-tree/
# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# @param root, a tree node
# @return a boolean
def isValidBST(self, root):
ans = True
pre,cur = None,root
while cur is not None:
if cur.left is None:
pre = cur
cur = cur.right
else:
ptr = cur.left
while ptr.right != None and ptr.right != cur:
ptr = ptr.right
if ptr.right == None:
ptr.right = cur
cur = cur.left
else:
ptr.right = None
pre = cur
cur = cur.right
if pre != None and cur != None:
if pre.val >= cur.val:
ans = False
return ansMorris Traversal: 非递归不用栈实现对树的中序遍历
原文地址：http://blog.csdn.net/tanzhangwen/article/details/40375029
