<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;

const int N = 30010;

int h[N], e[N << 1], ne[N << 1], idx;

int sz[N];
int down[N], up[N];

void add(int a, int b) {
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

pair<int, int> dfs1(int u, int p) {
    sz[u] = 1, down[u] = 0;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j != p) {
            auto t = dfs1(j, u);
            sz[u] += t.first;
            down[u] += t.second + t.first;
        }
    }
    return {sz[u], down[u]};
}

void dfs2(int u, int p, int n) {
    up[u] = 0;
    if (p != -1) {
//        up[u] = (up[p] + n - sz[p] + 1) + (down[p] - down[u] - 1 - sz[u] + sz[p] - sz[u]);
        up[u] = up[p] + n + down[p] - down[u] - sz[u] * 2;
    }
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (j != p) {
            dfs2(j, u, n);
        }
    }
}

class Solution {
public:
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>> &edges) {
        memset(h, -1, sizeof(int) * n), idx = 0;
        for (auto &edge: edges) {
            add(edge[0], edge[1]);
            add(edge[1], edge[0]);
        }
        dfs1(0, -1);
        dfs2(0, -1, n);
        vector<int> res(n);
        for (int i = 0; i < n; i++) {
            res[i] = up[i] + down[i];
        }
        return res;
    }
};
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