There is an undirected connected tree withnnodes labeled from0ton - 1andn - 1edges.You are given the integernand the arrayedgeswhereedges[i] = [ai, bi]indicates that there is an edge between nodesaiandbiin the tree.Return an arrayanswerof lengthnwhereanswer[i]is the sum of the distances between theithnode in the tree and all other nodes. 　　Example 1:

　　Input: n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]] Output: [8,12,6,10,10,10] Explanation: 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. 　　

class Solution {
public:
    vector<int> sumOfDistancesInTree(int
 
 n, vector<vector<int>>& edges) {
        //好久沒做和圖論相關的題目了 圖論的題目 dfs bfs？
        //連結圖連結串列
        vector<int> res(n),count(n);
        vector<vector<int>> tree(n);
        for(auto &edge:edges)
        {
            tree[edge[0]].push_back(edge[1]);
            tree[edge[
 
1]].push_back(edge[0]);
            
        }
        helper(tree,0,-1,count,res);
        helper2(tree,0,-1,count,res);
        return res;
        
    }
    
    void helper(vector<vector<int>> &tree,int cur,int pre,vector<int>&count,vector<int>& res)
    {
        
 
for(int i:tree[cur])
        {
            if(i==pre) continue;
            helper(tree,i,cur,count,res);
            count[cur]+=count[i];//這個統計更新的邏輯 還是不懂。。
            res[cur]+=res[i]+count[i];
        }
        ++count[cur];
    }
    void helper2(vector<vector<int>> &tree,int cur,int pre,vector<int>&count,vector<int>& res)
    {
        for(int i:tree[cur])
        {
            if(i==pre) continue;
            res[i]=res[cur]-count[i]+count.size()-count[i];
            helper2(tree, i, cur, count, res);
        }
    }
};