<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:
    // unordered_map<int, unordered_map<int, pair<int, int>>> pos_child;
    vector<vector<vector<int>>> pos_childNumDis;
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        pos_childNumDis.resize(n);
        vector<vector<int>> adj(n);
        for (auto &e : edges){
            adj[e[0]].push_back(e[1]);
            adj[e[1]].push_back(e[0]);
        }
        
        DFS_get_child(adj, -1, 0);
        vector<int> res(n, -1);
        DFS_res(adj, -1, 0, {0, 0}, res);
        return res;
    }
    pair<int, int> DFS_get_child(vector<vector<int>> &adj, int par, int root){
        pair<int, int> num_dis = {0, 0};
        for (int v : adj[root]){
            if (v == par)
                continue;
            pair<int, int> ret = DFS_get_child(adj, root, v);
            // pos_child[root][v] = ret;
            vector<int> vec{v, ret.first, ret.second};
            pos_childNumDis[root].push_back(vec);
            num_dis.first += ret.first;
            num_dis.second += ret.second;
        }
        num_dis.first++;
        num_dis.second += num_dis.first;
        return num_dis;
    }
    void DFS_res(vector<vector<int>> &adj, int par, int u, pair<int, int> parNumDis, vector<int> &res){
        pair<int, int> num_dis = parNumDis;
        // for (auto [k, p] : pos_child[u]){
        for (vector<int> &childNumDis : pos_childNumDis[u]){
            num_dis.first += childNumDis[1];
            num_dis.second += childNumDis[2];
        }
        res[u] = num_dis.second;

        num_dis.first++;
        // for (int v : adj[u]){
        for (vector<int> &childNumDis : pos_childNumDis[u]){
            int v = childNumDis[0];
            if (v == par)
                continue;
            
            num_dis.first -= childNumDis[1];
            num_dis.second -= childNumDis[2];

            DFS_res(adj, u, v, {num_dis.first, num_dis.second + num_dis.first}, res);

            num_dis.first += childNumDis[1];
            num_dis.second += childNumDis[2];
        }
    }
 
};
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