<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;
typedef long long ll;
#define minPq  priority_queue <ll, vector<ll>, greater<ll>> 



class Solution {
public:
    vector<int>dist,nodec,ans;
    vector<ll>filld(vector<vector<int>>&al,ll node,ll p)//dist,node
    {
        ll d=0;ll n=0;
        for(auto x:al[node])
        {
          if(x==p)continue;
          vector<ll>temp=filld(al,x,node);
          d+=temp[0];n+=temp[1];
        }
        
        dist[node]=d+n,nodec[node]=n+1;
        return {d+n,n+1};
    }
    void fillAns(vector<vector<int>>&al,ll node,ll p,ll sumd)
    {
      //base case
      //cal
      int n=al.size();
      ll remd=n-nodec[node];
      ans[node]=dist[node]+remd+sumd;
      for(auto x:al[node])
      {
          if(x==p)continue;
          fillAns(al,x,node,ans[node]-dist[x]-nodec[x]);
      }

    }
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        vector<vector<int>>al(n);
        for(auto x:edges)
        {
          al[x[0]].push_back(x[1]);
          al[x[1]].push_back(x[0]);
        }
        dist.resize(n+1);
        nodec.resize(n+1);
        ans.resize(n);
        filld(al,0,-1);
        fillAns(al,0,-1,0);
       
      return ans;
    }
};
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