简单换根DP

其实就是看好变化量，然后让父亲更新儿子就好了～

上图2当根节点的时候，ans[2] = ans[0] -sz[2]+n-sz[2];

class Solution {
public:vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {vector<vector<int>>g(n);for(auto &edge:edges){int a = edge[0],b = edge[1];g[a].push_back(b),g[b].push_back(a);}using ll = long long;vector<int>ans(n);vector<ll>sz(n+10);function<void(int,int,int)>dfs=[&](int u,int father,int dist){sz[u] = 1;ans[0]+=dist;for(auto &v:g[u]){if(v==father)continue;dfs(v,u,dist+1);sz[u]+=sz[v];}};function<void(int,int)>down=[&](int u,int father){for(auto &v:g[u]){if(v==father)continue;ans[v] = ans[u]-sz[v]+n-sz[v];down(v,u);}};dfs(0,-1,0);down(0,-1);return ans;}
};