剑指offer 27二叉搜索树与双向链表

class Solution {
public:
    void ConvertNode(TreeNode* pRootOfTree,TreeNode** pre)
    {
        if(pRootOfTree)
        {
            TreeNode* cur=pRootOfTree;
            if(cur->left)
                ConvertNode(cur->left,pre);
            cur->left=*pre;
            if(*pre)
                (*pre)->right=cur;
            *pre=cur;
            if(cur->right)
                ConvertNode(cur->right,pre);
        }
        
    }
    
    TreeNode* Convert(TreeNode* pRootOfTree)
    {
        if(!pRootOfTree)
            return NULL;
        TreeNode* pre=NULL;
        ConvertNode(pRootOfTree,&pre);
        while(pre->left)
        {
            pre=pre->left;
        }
        return pre;
    }
};