<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;
    long root_result;  //isme sirf root ka result store karunga starting mei

    vector<int> count;//isme mai count rakhunga kiska subtrees ka sabka
    //ye ek hi baari me hi calculate hojayega

    int dfs_base(unordered_map<int,vector<int>> &adj,int u,int parent,int curr_depth)
    {
        int total_count=1;

        root_result+=curr_depth;

        for(int v:adj[u])
        {
            if(v!=parent)
            {
                total_count+=dfs_base(adj,v,u,curr_depth+1);

            }
        }

        count[u]=total_count;
        return total_count;
    }

    // iske baad base node ka ans store hogya root_result mai


    //ab next dfs poora result vector ko populate karne ke lie

    void DFS(unordered_map<int,vector<int>> &adj,int u,int parent,vector<int> &result)
    {
        for(int v:adj[u])
        {
            if(v!=parent)
            {
                result[v]=result[u] -count[v]+(N-count[v]);
                //jo ki hamne formula derive kiya hai parent se baaki nodes ko populate karne ka
                //result parent ka - count child ka +remaining nodes 
                //remaining nodes hamare totalnodes- count of child se aajayenge



                 DFS(adj,v,u,result);
            }
        }

       

    } 

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {

        N=n;

        count.resize(N,0);

        //ab adj list bana lete jisko graph banana bhi kehte

        unordered_map<int,vector<int>> adj;

        for(auto vec:edges)
        {
            int u = vec[0];
            int v = vec[1];
            adj[u].push_back(v);
            adj[v].push_back(u);

        }
        root_result=0;

        dfs_base(adj,0,-1,0); //isme maine graph pass kiya
                             //fir source 0 aur parent -1
                             //aur alst 0 wo depth bataraha current

        vector<int> result(N,0);

        result[0]=root_result;

        DFS(adj,0,-1,result);

        return result;

        
    }
};
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