解题思路

从局部看整体，二叉搜索树中序有序，只要从中序遍历中转换就可以，完成一个结点得转换，再不断递归完成整颗树得转换

class Solution {
public:
    TreeNode* prev = NULL;
    TreeNode* first;
    void fun(TreeNode* node)
    {
        node->left = prev;
        if(prev!=NULL)
            prev->right = node;
        else 
            first = node;
        prev = node;
    }
    void Inorder(TreeNode* root)
    {
        if(root)
        {
            Inorder(root->left);
            fun(root);
            Inorder(root->right);
        }
    }
    TreeNode* convert(TreeNode* root) {
        first = NULL;
        Inorder(root);
        return first;
    }
};