<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>

#include <bits/stdc++.h>
using namespace std;

#define loop(i, a, n) for (lli i = (a); i < (n); ++i)
#define loopD(i, a, n) for (lli i = (a); i >= (n); --i)
#define all(c) (c).begin(), (c).end()
#define rall(c) (c).rbegin(), (c).rend()
#define sz(a) ((int)a.size())
#define YES cout << "YES" << endl;
#define NO cout << "NO" << endl;
#define endl '\n'
#define fastio std::ios::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
#define pb push_back
#define pp pop_back()
#define fi first
#define si second
#define v(a) vector<int>(a)
#define vv(a) vector<vector<int>>(a)
#define present(c, x) ((c).find(x) != (c).end())
#define set_bits __builtin_popcountll
#define MOD 1000000007
// #define int long long

typedef long long lli;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<lli, lli> pll;
typedef pair<int, int> pii;
typedef unordered_map<int, int> umpi;
typedef map<int, int> mpi;
typedef vector<pii> vp;
typedef vector<lli> vll;
typedef vector<vll> vvll;

class Solution
{
public:
    static const int m = 3 * 1e4 + 10;
    vi g[m];
    int cnt[m], subAns[m];
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>> &edges)
    {
        fastio;
        for (const auto &e : edges)
        {
            g[e[0]].pb(e[1]);
            g[e[1]].pb(e[0]);
        }
        vi ans(n);
        auto add = [&](int node, int child)
        {
            cnt[node] += cnt[child];
            subAns[node] += subAns[child] + cnt[child];
        };

        auto subtract = [&](int node, int child)
        {
            cnt[node] -= cnt[child];
            subAns[node] -= subAns[child] + cnt[child];
        };

        auto changeRoot = [&](int node, int child)
        {
            subtract(node, child);
            add(child, node);
        };

        function<void(int, int)> dfs = [&](int node, int parent) -> void
        {
            subAns[node] = 0;
            cnt[node] = 1;
            for (auto child : g[node])
            {
                if (child == parent)
                    continue;
                dfs(child, node);
                add(node, child);
            }
        };

        function<void(int, int)> reRoot = [&](int node, int parent) -> void
        {
            // initialize ans node
            ans[node] = subAns[node];
            for (int child : g[node])
            {
                if (child == parent)
                    continue;
                changeRoot(node, child);
                reRoot(child, node);
                changeRoot(child, node); // backtrack
            }
        };
        dfs(0, -1);
        reRoot(0, -1);
        return ans;
    }
};
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