<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>

pair<int,int> dfs1(int i,unordered_map<int,vector<int>> &adjlist,vector<int> &s,vector<int> &x,vector<int> &visited,vector<int> &parent){
          if(s[i] != -1 && x[i]!=-1){
            return make_pair(s[i],x[i]);
          }
          int sum = 0;
          int num = 1;
          for(auto neigh : adjlist[i]){
            if(!visited[neigh]){
                parent[neigh]=i;
                visited[neigh]=1;
                pair<int,int> adj = dfs1(neigh,adjlist,s,x,visited,parent);
                num = num + adj.second;
                sum = sum + adj.first+adj.second;
            }
          }
          s[i]=sum;
          x[i]=num;
          return make_pair(sum,num);
}
void dfs2(int i,unordered_map<int,vector<int>> &adjlist,vector<int> &s,vector<int> &x,vector<int> &visited1,vector<int> &l,vector<int> &parent){
          int n = s.size();
          int par = parent[i];
          if(par == -1){
            l[i]=0;
          }
          else{
             l[i]= n + l[par]+s[par]-s[i]-2*x[i];
          }
          for(auto neigh : adjlist[i]){
            if(!visited1[neigh]){
                visited1[neigh]=1;
                dfs2(neigh,adjlist,s,x,visited1,l,parent);
            }
          }
          return;
}
class Solution {
public:
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        int m = edges.size();
        unordered_map<int,vector<int>> adjlist;
        for(int i=0;i<m;i++){
            adjlist[edges[i][0]].push_back(edges[i][1]);
            adjlist[edges[i][1]].push_back(edges[i][0]);
        }
        vector<int> s(n,-1);
        vector<int> visited(n,0);
        vector<int> parent(n,-1);
        vector<int> x(n,-1);
        visited[0]=1;
        pair<int,int> temp = dfs1(0,adjlist,s,x,visited,parent);
        vector<int> l(n,0);
        vector<int> visited1(n,0);
        visited1[0]=1;
        dfs2(0,adjlist,s,x,visited1,l,parent);
        vector<int> ans(n,0);
        for(int i=0;i<n;i++){
            ans[i]=s[i]+l[i];
        }
        return ans;

    }
};
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