sum of Distances in Tree solution code example

Example 1: Sum of Distances in Tree

int[] res, count;
    ArrayList<HashSet<Integer>> tree;
    public int[] sumOfDistancesInTree(int N, int[][] edges) {
        tree = new ArrayList<HashSet<Integer>>();
        res = new int[N];
        count = new int[N];
        for (int i = 0; i < N ; ++i)
            tree.add(new HashSet<Integer>());
        for (int[] e : edges) {
            tree.get(e[0]).add(e[1]);
            tree.get(e[1]).add(e[0]);
        }
        dfs(0, -1);
        dfs2(0, -1);
        return res;
    }

    public void dfs(int root, int pre) {
        for (int i : tree.get(root)) {
            if (i == pre) continue;
            dfs(i, root);
            count[root] += count[i];
            res[root] += res[i] + count[i];
        }
        count[root]++;
    }


    public void dfs2(int root, int pre) {
        for (int i : tree.get(root)) {
            if (i == pre) continue;
            res[i] = res[root] - count[i] + count.length - count[i];
            dfs2(i, root);
        }
    }

Example 2: Sum of Distances in Tree

vector<unordered_set<int>> tree;
    vector<int> res, count;

    vector<int> sumOfDistancesInTree(int N, vector<vector<int>>& edges) {
        tree.resize(N);
        res.assign(N, 0);
        count.assign(N, 1);
        for (auto e : edges) {
            tree[e[0]].insert(e[1]);
            tree[e[1]].insert(e[0]);
        }
        dfs(0, -1);
        dfs2(0, -1);
        return res;

    }

    void dfs(int root, int pre) {
        for (auto i : tree[root]) {
            if (i == pre) continue;
            dfs(i, root);
            count[root] += count[i];
            res[root] += res[i] + count[i];
        }
    }

    void dfs2(int root, int pre) {
        for (auto i : tree[root]) {
            if (i == pre) continue;
            res[i] = res[root] - count[i] + count.size() - count[i];
            dfs2(i, root);
        }
    }