题解 | #二叉搜索树与双向链表#

class Solution 
{
public:
    TreeNode* Convert(TreeNode* pRootOfTree) 
    {
        if(pRootOfTree == NULL) return NULL;
        //cout<< pRootOfTree->val <<' ';
        TreeNode* l = Convert(pRootOfTree->left); //l是左子树转换后的最左边结点
        TreeNode *ans = l;
        TreeNode* r = Convert(pRootOfTree->right);
//r是右子树转换后的最左边结点
        if(l != NULL)

        {
            while(l->right != NULL) l=l->right;
            l->right = pRootOfTree;
        }
        pRootOfTree ->left = l;
        if( r!= NULL) r->left = pRootOfTree;
        pRootOfTree -> right = r;
        if(ans == NULL) ans = pRootOfTree;
        return ans;
    }
};