A clean solution in C (time complexity O(n^3)) | ☺
Complexity

Time Complexity: O(n³) - This comes from the nested DFS traversals and combination calculations for each node and edge.


Space Complexity: O(n²) -
## Complexity
- Time Complexity: O(n³) - This comes from the nested DFS traversals and combination calculations for each node and edge.

- Space Complexity: O(n²) - We store intermediate results for subtree combinations during the DFS.

## Code Explanation

1. Tree Construction: Builds the adjacency list from input edges.

1. DFS Function: Recursively explores subtrees and calculates the number of nodes at each depth.

1. Combination Function: Merges subtree counts from different child branches.

1. Result Calculation: Aggregates counts for all possible diameters by considering both edge-centered and node-centered cases.

# Code
```c []

#define MAX_N 15
#define min(a, b) ((a) < (b) ? (a) : (b))
typedef struct 
{
    int size;
    int data[MAX_N];
} array_t;


array_t __adj[MAX_N];


array_t *combinations(array_t *arr1, array_t *arr2) 
{
    if (arr1->size > arr2->size)
        return combinations(arr2, arr1);
    
    array_t *ret = calloc(1, sizeof(array_t));
    ret->size = arr2->size;

    ret->data[0] = arr1->data[0] * arr2->data[0];
    
    for (int i = 1; i < arr1->size; ++i) 
        arr1->data[i] += arr1->data[i - 1];

    for (int i = 1; i < arr2->size; ++i) 
        arr2->data[i] += arr2->data[i - 1];

    for (int i = 1; i < arr1->size; ++i)
        ret->data[i] = arr1->data[i] * arr2->data[i] - arr1->data[i - 1] * arr2->data[i - 1];
    
    for (int i = arr1->size; i < arr2->size; ++i)
        ret->data[i] = (arr2->data[i] - arr2->data[i - 1]) * arr1->data[arr1->size - 1];
    

    return ret;
}

array_t *dfs(int root, int parent) 
{
    array_t *ret = malloc(sizeof(array_t));
    ret->data[0] = 1;
    ret->size = 1;
    
    array_t *tmp, *combined;
    for (int i = 0; i < __adj[root].size; ++i) 
    {
        int node = __adj[root].data[i];
        if (node == parent) 
            continue;
        
        tmp = dfs(node, root);

        combined = combinations(tmp, ret);
        free(tmp);
        free(ret);

        ret = combined;
    }
    
    memmove(ret->data + 1, ret->data, ret->size * sizeof(int));
    ret->data[0] = 1;
    ++ret->size;

    return ret;
}

int *countSubgraphsForEachDiameter(int n, int** edges, int edges_size, int *, int *ret_size) 
{
    for (int i = 0; i < n; ++i)
        __adj[i].size = 0;
    
    for (int i = 0; i < edges_size; ++i) 
    {
        int u = edges[i][0] - 1, 
            v = edges[i][1] - 1;
        __adj[u].data[__adj[u].size++] = v;
        __adj[v].data[__adj[v].size++] = u;
    }
    
    int ans[MAX_N] = { 0 };

    for (int i = 0; i < edges_size; ++i) 
    {
        int u = edges[i][0] - 1, 
            v = edges[i][1] - 1;
        
        array_t *u_tree = dfs(u, v),
                *v_tree = dfs(v, u);

        int min_len = min(u_tree->size, v_tree->size);
        for (int j = 1; j < min_len; ++j)
            ans[(j << 1) - 1] += u_tree->data[j] * v_tree->data[j];

        free(u_tree);
        free(v_tree);
    }
    
    for (int root = 0; root < n; ++root) 
    {
        if (__adj[root].size == 1) 
            continue; 
        
        array_t *tree = dfs(__adj[root].data[0], root);
        
        int curr[MAX_N] = { 0 };
        
        for (int i = 1; i < __adj[root].size; ++i) 
        {
            array_t *cur_tree = dfs(__adj[root].data[i], root);

            int j = 1, prefix = cur_tree->data[1] + 1;
            for (int k = 4; k < n; k += 2) 
            {
                while (j + 1 < min(k >> 1, cur_tree->size))
                    prefix += cur_tree->data[++j];

                curr[k] *= prefix;
            }
            
            int min_len = min(tree->size, cur_tree->size);
            for (int k = 0; k < min_len; ++k) 
                curr[k << 1] += tree->data[k] * cur_tree->data[k];
        
            
            array_t *tmp = combinations(tree, cur_tree);
            
            free(tree);
            free(cur_tree);

            tree = tmp;
        }
        
        free(tree);

        for (int k = 2; k < n; k += 2)
            ans[k] += curr[k];
    }


    int *ret = malloc((n - 1) * sizeof(int));
    *ret_size = n - 1;

    memcpy(ret, ans + 1, (n - 1) * sizeof(int));
    

    return ret;
}
```

## Attribution
This [solution](https://leetcode.com/problems/count-subtrees-with-max-distance-between-cities/solutions/3651372/o-n-3-algorithm) was inspired by the approach shared by user **bangyewu**. 
['C']