<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>

class Solution {
public:
int N;
vector<int>sum;
int root_sum;
vector<int>count;
vector<int>parent;

    int dfs1(int s, int dis, unordered_map<int, vector<int>>&mp,vector<int>&vis){
         vis[s]=1;

         sum[0]+=dis;
         int temp=0;
         for(auto x: mp[s]){
 
            if(!vis[x]){
               temp+=dfs1(x,dis+1,mp,vis);
            }
         }
         temp++;
         count[s]=temp;
         return temp;
    }

    void dfs2(int s, unordered_map<int, vector<int>>& mp, vector<int>& vis) {
        vis[s] = 1;
        for (auto x : mp[s]) {
            if (!vis[x]) {
                sum[x] = sum[s] - count[x] + (N - count[x]);
                dfs2(x, mp, vis);
            }
        }
    }


    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        unordered_map<int, vector<int>>mp;
        N=n;
        root_sum=0;
        count.resize(n);
        sum.resize(n,0);
        parent.resize(n,0);
        for(auto x:edges){
            mp[x[0]].push_back(x[1]);
            parent[x[1]]=x[0];
            mp[x[1]].push_back(x[0]);
        }
        vector<int>vis(n,0);
        dfs1(0,0,mp,vis);// to calculate 0 ka distance sabse and cout of below nodes (for all nodes-->top-bottom approach kaam nahi aai to bottom-up use kari)

        // for(int i=1;i<n;i++){
        //     int temp=sum[parent[i]];
        //     sum[i]+=temp-count[i] + (n-count[i]);
        // }

        fill(vis.begin(), vis.end(), 0);

        // Second DFS to update distances based on the first DFS results
        dfs2(0, mp, vis);
     
        return sum;
        
      

    }
};
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