问题描述：

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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

基本思路：

二叉树中序便利。最简单的递归写法。

代码：

/** 
 * Definition for binary tree 
 * struct TreeNode { 
 *     int val; 
 *     TreeNode *left; 
 *     TreeNode *right; 
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 
 * }; 
 */
void myinorderTraversal(TreeNode * root,vector<int>&result)  //C++
    {
        if(root == NULL)
            return;
        if(root->left != NULL)
            myinorderTraversal(root->left,result);
        result.push_back(root->val);
        if(root->right != NULL)
            myinorderTraversal(root->right,result);
            
    }
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> result;
        if(root == NULL)
            return result;
        
        myinorderTraversal(root,result);
        return result;
    }

[leetcode]

原文：http://blog.csdn.net/chenlei0630/article/details/41523673