<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>

#define ll long long

class Solution {
public:
    int N;
    vector<int> count_child;
    ll root_result;

    int first_dfs(unordered_map<int, vector<int>> &adj, int node, int parent, int curr_depth){
        int total_count = 1; // count of child including root itself, this 1 is for the root;

        root_result += curr_depth;

        for(auto& it: adj[node]){
            if(it == parent) continue;

            total_count += first_dfs(adj, it, node, curr_depth+1);
        }

        count_child[node] = total_count;

        return total_count;
    }


    void final_dfs(int node, int parent, unordered_map<int, vector<int>> &adj, vector<int>& dist){
        for(int &it : adj[node]){
            if(it == parent) continue;

            dist[it] = dist[node] - count_child[it] + (N - count_child[it]);

            final_dfs(it, node, adj, dist);
        }

    }

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        N = n;
        count_child.resize(N, 0);

        unordered_map<int, vector<int>> adj;
        for(auto it: edges){
            adj[it[0]].push_back(it[1]);
            adj[it[1]].push_back(it[0]);
        }
        root_result = 0;
        first_dfs(adj, 0, -1, 0);

        vector<int> dist(n, 0);
        dist[0] = root_result;

        final_dfs(0, -1, adj, dist);

        return dist;
        // --------------BFS---------------------

        // queue<pair<int, int>> q;
        // q.push({0, 0});

        // vector<int> vis(n, 0);

        // while(!q.empty()){
        //     int size = q.size();
        //     while(size--){
        //         auto it = q.front();
        //         int node = it.first;
        //         int dist = it.second;
        //         q.pop();

        //         vis[node] = 1;
        //         sum0 += dist;

        //         for(auto i: adj[node]){
        //             if(vis[i] == 0){
        //                 q.push({i, dist+1});
        //             }
        //         }
        //     }
        // }

        // --------------X------------------------
    
    }
};
121906