We were discussing Master Data management and how it differs between MongoDB and traditional databases.
Master Data Management participates in:
Planning and architecture
Execution and deployment
Quality and monitoring
Governance and Conrol
Standards and Compliance

Due to the lagging embrace of cloud based master data management technologies ( with the possible exception of Snowflake due to its architecture ),  Riversand still continues to enjoy widespread popularity in an on-premise solution.
Riversand offers data modeling, data synchronization, data standardization, and flexible workflows within the tool. It offers scalability, performance and availability based on its comprehensive Product Information Management (PIM) web services. Functionality is provided in layers of information management starting with print/translation workflows at the bottom layer, followed by workflow or security for access to the assets, their editing, insertions and bulk insertions, followed by integrations or portals  followed by flexible integration capabilities, full/data exports, multiple exports, followed by integration portals for integration with imports/exports, data pools and platforms, followed by digital asset management layer for asset on-boarding and delivery to channels, and lastly data management for searches, saved searches, channel based content, or localized content and the ability to author variants, categories, attributes and relationships to stored assets.



#codingexercise
Find if one binary tree is a subtree of another
We were discussing recursive solution and why it does not apply when the subtree is not a leaf in the original. There is also an iterative solution that can be modified to suit this purpose.
bool IsEqualIterative(Node root1, Node root2, ref Stack<Node> stk)
{
var current1 = root1;
var current2 = root2;

while (current1 != null ||  current2 != null || stk. empty() == false)
{

if (current1 == null && current2 != null ) return false;
if (current1 != null && current2 == null ) return false;

if (current1 != null)
{
if (current1.data != current2.data) return false;
stk.push(current1);
current1 = current1.left;
current2 = current2.left;
continue;
}

if (stk. empty() == false)
{
// we can introduce sequence equal-logic-so-far here
current1 = stk. pop();
current1 = current1.right;
current2 = current2.right;
}

}
return true;
}