49. 二叉搜索树与双向链表

 https://www.bilibili.com/video/av91129831

栈做https://www.nowcoder.com/profile/6084631/codeBookDetail?submissionId=15463248

队列做https://www.nowcoder.com/profile/1763495/codeBookDetail?submissionId=16862302

数组：https://www.nowcoder.com/profile/5530542/codeBookDetail?submissionId=14998267

class Solution {
public:
TreeNode* convert(TreeNode* root)
{

        if(!root) return NULL;
        vector<TreeNode* > vec;
        cout<<"&vec="<<&vec<<endl;
        dfs(root,vec);
        for(int i =0;i < vec.size()-1;i++)
        {
            vec[i]->right=vec[i+1];
            vec[i+1]->left=vec[i];
        }
         vec[0]->left=vec[vec.size()-1]->right=NULL;
        return vec[0];
            
    }
    
    void dfs(TreeNode* root,vector<TreeNode* >  &vec)
    {
        cout<<"&vec2="<<&vec<<endl;
        if(!root) return;
        dfs(root->left,vec);
        
        vec.push_back(root);
        
        dfs(root->right,vec);
        
    }

};

 可以看到convert(TreeNode* root) 中的

 vector<TreeNode* > vec;和

 void dfs(TreeNode* root,vector<TreeNode* >  &vec)的

 &vec地址相同