<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>

using vi = vector<int>;
using vvi = vector<vi>;
using int2 = pair<int, int>;
class Solution {
public:
    vvi adj;
    vi dp;
    vi sizes;
    vi ans;
    int total;

    int2 dfs(int node, int p){

        for(auto nb: adj[node]){
            if(nb == p) continue;

            auto [ans, sz] = dfs(nb, node);

            sizes[node] += sz;
            dp[node] += sz + ans;
        }

        return {dp[node], sizes[node]};        

    }

    void calc(int node, int p){

        for(auto nb: adj[node]){
            if(nb == p) continue;

            ans[nb] = ans[node] + total - 2 * sizes[nb];

            calc(nb, node);
        }

    }

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges){
        adj.resize(n);

        for(auto e: edges){
            adj[e[0]].push_back(e[1]);
            adj[e[1]].push_back(e[0]);
        }

        dp.assign(n, 0);

        sizes.assign(n, 1);

        dfs(0, -1);

        // for(int i=0; i<n; i++) cout<<dp[i]<<" ";
        // cout<<"\n";

        // for(int i=0; i<n; i++) cout<<sizes[i]<<" ";
        // cout<<"\n";

        ans.assign(n, 0);
        ans[0] = dp[0];

        total = sizes[0];

        ans[0] = dp[0];

        calc(0, -1);

        return ans;

        
    }
};
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