題目描述：
輸入一棵二叉搜索樹，將該二叉搜索樹轉換成一個排序的雙向鏈表。要求不能創(chuàng)建任何新的結點，只能調整樹中結點指針的指向。
關鍵點：1.遞歸
2.遞歸過程中鏈接左右指針

class Solution {
public:
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(pRootOfTree == NULL) return pRootOfTree;
        pRootOfTree = ConvertNode(pRootOfTree);
        while(pRootOfTree->left) pRootOfTree = pRootOfTree->left;
        return pRootOfTree;
    }
 
    TreeNode* ConvertNode(TreeNode* root)
    {
        if(root == NULL) return root;
        if(root->left)
        {
            TreeNode *left = ConvertNode(root->left);
            while(left->right) left = left->right;
            left->right = root;
            root->left = left;
        }
         
        if(root->right)
        {
            TreeNode *right = ConvertNode(root->right);
            while(right->left) right = right->left;
            right->left = root;
            root->right = right;
        }
        return root;
    }
};