Need help recursion explanation Leetcode

Question

How does this code work? (leetcode 95 question) I don't understand how the 2 recursions work inside the for loop. Does the 2nd inner for loop end when the recursive function returns NULL? Or would it continue executing the 3rd inner for loop?

class Solution {
public:
    vector<TreeNode*> generateTrees(int n) {
        
        
        if(n == 0) {
            return {};
        }
        
        vector<TreeNode*> ans = generateT(1,n);
        return ans;
        
        
    }
    
    
    vector<TreeNode*> generateT(int l, int r) {
        if(l > r) return {nullptr};
        vector<TreeNode*> ans;
        for(int i=l; i <= r; ++i) {
            for(TreeNode*left: generateT(l, i-1)) {
                for(TreeNode* right:generateT(i+1, r)) {
                    ans.push_back(new TreeNode(i));
                    ans.back()->left = left;
                    ans.back()->right = right;
                }
            }
        }
        return ans;
    }
};

Problem statement: Given an integer n, return all the structurally unique BST's (binary search trees), which has exactly n nodes of unique values from 1 to n. Return the answer in any order.

Answer 1

Does the 2nd inner for loop end when the recursive function returns NULL?

No. The recursive function is not returing NULL, it is returning vector of nullptr.

Or would it continue executing the 3rd inner for loop?

Of course, it will.

How does this code work? I don't understand how the 2 recursions work inside the loop.

I suppose the following snippet is the cause of confusion, so commented the case when nullptr provided by outer loop.

 vector<TreeNode*> generateT(int l, int r) {
        
        if(l > r) return { nullptr };
        vector<TreeNode*> ans;
        
        for ( int i = l; i <= r; i++ ) {
            // if l = 0, i = 0
            for ( TreeNode* left :generateT(l, i-1) ) // if l = 0, i = -1, returns { nullptr } (vector of nullptr)
                for (TreeNode* right :generateT(i+1, r)) { // now this snippet will execute
                    auto node = new TreeNode(i);
                    ans.push_back(node);
                    node->left = left;  // the nullptr we have from the outer loop, will provide null value for this
                    node->right = right;
                }
        }
            
        return ans;
    }

Visually, for a combination of node where,

  a
   \
    b
   / \
null  c
     /
    null

the above pattern occurs the provided { nullptr } from outer loop will come in handy setting left node.