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].

解法一：递归

1 /** 2  * Definition for a binary tree node. 3  * struct TreeNode { 4  *     int val; 5  *     TreeNode *left; 6  *     TreeNode *right; 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8  * }; 9  */10 class Solution {11 public:12     vector
     
       inorderTraversal(TreeNode* root) {13         vector
      
        result;14         inorder(root,result);15         return result;       16         17     }18     19     void inorder(TreeNode *node,vector
       
         &result)20     {21         if(node!=NULL)22         {23             inorder(node->left,result);24             result.push_back(node->val);25             inorder(node->right,result);26         }27         28     }29 };