题目：

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

OJ‘s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.

Here‘s an example:

   1
  /  2   3
    /
   4
         5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

提示：

题目要求给出二叉树的中序遍历，总共有三种方法可以使用：

• 函数递归
• 利用Stack迭代
• Morris遍历法

其中递归法比较简单，这里就不赘述了。下面的代码部分主要贴出第二和第三种方法，其中关于Morris遍历法的解释，可以点击链接查看。

代码：

首先是利用Stack迭代：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> result;
        stack<TreeNode *> stack;
        TreeNode *pCurrent = root;

        while(!stack.empty() || pCurrent)
        {
            if(pCurrent)
            {
                stack.push(pCurrent);
                pCurrent = pCurrent->left;
            }
            else
            {
                TreeNode *pNode = stack.top();
                result.push_back(pNode->val);
                stack.pop();
                pCurrent = pNode->right;
            }
        }
        return result;
    }
};

然后是使用Morris遍历：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        TreeNode *cur = root, *prev = NULL;
        while (cur != NULL) {
            if (cur->left == NULL) {
                result.push_back(cur->val);
                cur = cur->right;
            } else {
                prev = cur->left;
                while(prev->right != NULL && prev->right != cur)
                    prev = prev->right;

                if (prev->right == NULL) {
                    prev->right = cur;
                    cur = cur->left;
                } else {
                    prev->right = NULL;
                    result.push_back(cur->val);
                    cur = cur->right;
                }
            }
        }
        return result;
    }
};