Another approach

#1

I tried this question using queue

Is this approach also right for this question ?

#2

yes, it seems right, you can also do this

void flatten(TreeNode* &root) {
        if(root == nullptr) return;
        stack<TreeNode*>q;
        q.push(root);
        while(!q.empty()){
            TreeNode* curr = q.top();
            q.pop();
            if(curr->right){ 
                q.push(curr->right);
            }
            if(curr->left){ 
                q.push(curr->left);
            }
            if(!q.empty()){
                curr->right = q.top();
            }
            curr->left = nullptr;
        }
        
        return;
    }

this is for bfs approach.

#3

ooh yes yes, but these approaches require a extra space for the logic, while bhaiya’s approach just plays with pointers so i guess thats better.

#4

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