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

For example:

   1    \     2    /   3

return [1,3,2].

/** * 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
     
       inorderTraversal(TreeNode* root) {        vector
      
        vec;        inOrder(root, vec);        return vec;    }        /*    void inOrder(TreeNode *root, vector
       
         &path)    {        //递归写法        if (root)        {            inOrder(root->left, path);            path.push_back(root->val);            inOrder(root->right, path);        }    }    */    //非递归写法    void inOrder(TreeNode *root, vector
        
          &path)    {        stack
         
           TreeNodeStack;        while (root != NULL || !TreeNodeStack.empty())        {            while(root != NULL)            {                TreeNodeStack.push(root);                root = root->left;            }            if (!TreeNodeStack.empty())            {                root = TreeNodeStack.top();                TreeNodeStack.pop();                path.push_back(root->val);                root = root->right;            }        }    }};