16:I[6874,["874","static/chunks/874-a7ac5a00a7945b42.js","178","static/chunks/178-ab4037c28a81913e.js","844","static/chunks/844-c7a0e2759f5c4b8a.js","99","static/chunks/app/topics/dsa/%5BsubTopicSlug%5D/%5BpostSlug%5D/page-646502347b3d846f.js"],""] 17:I[9844,["874","static/chunks/874-a7ac5a00a7945b42.js","178","static/chunks/178-ab4037c28a81913e.js","844","static/chunks/844-c7a0e2759f5c4b8a.js","99","static/chunks/app/topics/dsa/%5BsubTopicSlug%5D/%5BpostSlug%5D/page-646502347b3d846f.js"],"default"] 18:T53f,

If you can't understand this, ngmi. But still here's how the recursion works:

     1
    / \
   2   3
  / \
 4   5

Start at root (1):

Calls findHeight(2) for left subtree
Calls findHeight(3) for right subtree

For node (2):
Calls findHeight(4) for left subtree
Calls findHeight(5) for right subtree

For node (4):
Left child is nullptr → returns 0
Right child is nullptr → returns 0
Takes max(0,0) + 1 = 1

For node (5):
Left child is nullptr → returns 0
Right child is nullptr → returns 0
Takes max(0,0) + 1 = 1

Back to node (2):
Left height = 1 (from node 4)
Right height = 1 (from node 5)
Takes max(1,1) + 1 = 2

For node (3):
Left child is nullptr → returns 0
Right child is nullptr → returns 0
Takes max(0,0) + 1 = 1

Finally, back to root (1):
Left height = 2 (from node 2)
Right height = 1 (from node 3)

Takes max(2,1) + 1 = 3

int findHeight(Node* root) {
    // Base case: if tree is empty
    if (root == nullptr) {
        return 0;
    }
    
    // Recursively find height of left and right subtrees
    int leftHeight = findHeight(root->left);
    int rightHeight = findHeight(root->right);
    
    // Return the maximum height plus 1 for current node
    return max(leftHeight, rightHeight) + 1;
}

8:["$","div",null,{"className":"post-page-gradient-background","children":["$","article",null,{"className":"container mx-auto px-4 py-12 md:py-16 pt-20","children":["$","div",null,{"className"