<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>

#pragma GCC optimize("O3", "unroll-loops")
int speedup = []{ios::sync_with_stdio(0); cin.tie(0);cout.tie(NULL); return 0;}();
//ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
class Solution {
public:
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        vector<int> gp[n];
        for (int i = 0; i < edges.size(); ++i) {
            gp[edges[i][0]].push_back(edges[i][1]);
            gp[edges[i][1]].push_back(edges[i][0]);
        }

        vector<int> dp1(n,0);
        vector<int> subtreeSize(n,1);
        vector<int> ans(n,0);
        auto dfs = [&](auto &self,int node, int pt)->void {
            for (const auto &child : gp[node]) {
                if (child == pt) continue;
                self(self,child,node);
                dp1[node] += subtreeSize[child] + dp1[child];
                subtreeSize[node] += subtreeSize[child];
            }
        };

        auto dfs2 = [&](auto &self, int node, int pt)->void {        
            for (const auto &child : gp[node]) {
                if (child == pt)continue;
                //ans for 2 would have be calculated when we calculate for zer0, so just remove the number of times zero has to be travelled,
                // this is achieved by removeing subtreesize
                //Second part is when need to add contribution from parent nodes, in this case parent subtree would contain nodes 0,1 and it should be added.
                ans[child] = ans[node] - subtreeSize[child] + (n - subtreeSize[child]);
                self(self,child,node); 
            }
        };

        dfs(dfs,0,-1);
        ans[0] = dp1[0];
        dfs2(dfs2,0,-1);
        return ans;
    }
};
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