<p>There is an undirected connected tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code> and <code>n - 1</code> edges.</p>

<p>You are given the integer <code>n</code> and the array <code>edges</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>

<p>Return an array <code>answer</code> of length <code>n</code> where <code>answer[i]</code> is the sum of the distances between the <code>i<sup>th</sup></code> node in the tree and all other nodes.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist1.jpg" style="width: 304px; height: 224px;" />
<pre>
<strong>Input:</strong> n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
<strong>Output:</strong> [8,12,6,10,10,10]
<strong>Explanation:</strong> The tree is shown above.
We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
equals 1 + 1 + 2 + 2 + 2 = 8.
Hence, answer[0] = 8, and so on.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist2.jpg" style="width: 64px; height: 65px;" />
<pre>
<strong>Input:</strong> n = 1, edges = []
<strong>Output:</strong> [0]
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/23/lc-sumdist3.jpg" style="width: 144px; height: 145px;" />
<pre>
<strong>Input:</strong> n = 2, edges = [[1,0]]
<strong>Output:</strong> [1,1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>edges.length == n - 1</code></li>
	<li><code>edges[i].length == 2</code></li>
	<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>
	<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>
	<li>The given input represents a valid tree.</li>
</ul>

class Solution {
public:
    // void distChildren(int root, int parent, vector<int>* graph,
    //                   vector<pair<int, int>>* dist) {
    //     (*dist)[root].second = 1;

    //     for (int i = 0; i < graph[root].size(); i++) {
    //         if (graph[root][i] != parent) {
    //             distChildren(graph[root][i], root, graph, dist);

    //             (*dist)[root].first += (*dist)[graph[root][i]].first +
    //                                    (*dist)[graph[root][i]].second;
    //             (*dist)[root].second += (*dist)[graph[root][i]].second;
    //         }
    //     }
    // }

    // void DFS1(int root, int parent, vector<int>* graph,
    //           vector<pair<int, int>>* dist, vector<int>* answer, int d,
    //           int cnt) {
    //     (*answer)[root] = d + (*dist)[root].first;

    //     for (int i = 0; i < graph[root].size(); i++) {
    //         if (graph[root][i] != parent) {
    //             int new_d = d + (*dist)[root].first -
    //                         (*dist)[graph[root][i]].first -
    //                         (*dist)[graph[root][i]].second;
    //             int new_cnt =
    //                 cnt + (*dist)[root].second -
    //                 (*dist)[graph[root][i]].second;

    //             DFS1(graph[root][i], root, graph, dist, answer, new_d +
    //             new_cnt,
    //                  new_cnt);
    //         }
    //     }
    // }

    // retrun Sum of Distances in subtree
    int DFS(int root, int parent, vector<int>* graph, vector<int>* cnt) {
        int sum = 0;

        for (int i = 0; i < graph[root].size(); i++) {
            int child = graph[root][i];

            if (child != parent) {
                sum += DFS(child, root, graph, cnt) + (*cnt)[child];

                (*cnt)[root] += (*cnt)[child];
            }
        }

        return sum;
    }

    void DFS1(int root, int parent, int n, vector<int>* graph, const vector<int>& cnt, vector<int>* answer) {
        for (int i = 0; i < graph[root].size(); i++) {
            int child = graph[root][i];

            if (child != parent) {
                 (*answer)[child] = (*answer)[root] - cnt[child] + (n - cnt[child]);

                DFS1(child, root, n, graph, cnt, answer);
            }
        }
    }

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        // adjacency list
        vector<int>* graph = new vector<int>[n];
        for (int i = 0; i < edges.size(); i++) {
            int u_i = edges[i][0];
            int v_i = edges[i][1];

            graph[u_i].push_back(v_i);
            graph[v_i].push_back(u_i);
        }

        vector<int> cnt(n, 1); // nodes in subtree
        vector<int> answer(n);

        // first DFS, calculate answer[0] and fill the cnt array
        answer[0] = DFS(0, -1, graph, &cnt);

        DFS1(0, -1, n, graph, cnt, &answer);

        // DFS1();

        // // dist, node cnt
        // distChildren(0, -1, graph, &dist);
        // DFS(0, -1, graph, &dist, &answer, 0, 0);

        return answer;
    }
};
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