<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 dfs1(int source, vector<int>& visited,
             unordered_map<int, list<int>> adj, int disTillNow) {

        visited[source] = 1;

        int currDis = disTillNow;

        for (auto nei : adj[source]) {
            if (!visited[nei]) {
                currDis = currDis + dfs1(nei, visited, adj, disTillNow + 1);
            }
        }
        return currDis;
    }

    int dfs2(unordered_map<int, list<int>>& adj, int u, int prev, int& root_result, int current_depth, vector<int>& count) {

        int total_count = 1;

        root_result += current_depth;

        for (auto& v : adj[u]) {
            if (v == prev)
                continue;

            total_count +=
                dfs2(adj, v, u, root_result, current_depth + 1, count);
        }

        count[u] = total_count;

        return total_count;
    }

    void dfs3(unordered_map<int, list<int>>& adj, int u, int parent, vector<int>& result, int n, vector<int>& count) {

        for (auto& v : adj[u]) {
            if (v == parent)
                continue;

            result[v] += result[u] - count[v] + (n - count[v]);
            dfs3(adj, v, u, result, n, count);
        }
    }

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

        /*
        unordered_map<int, list<int>> adj;
        for (auto i : edges) {
            int u = i[0];
            int v = i[1];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }

        vector<int> visited(n, 0);
        int dis = 0;
        vector<int> ans;
        for (int i = 0; i < n; i++) {
            vector<int> visited(n, 0);
            ans.push_back(dfs1(i, visited, adj, 0));
        }
        return ans;

        */

        unordered_map<int, list<int>> adj;
        for (auto i : edges) {
            int u = i[0];
            int v = i[1];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }

        int root_result = 0;
        vector<int> count(n, 0);
        vector<int> result(n, 0);

        dfs2(adj, 0, -1, root_result, 0, count);

        result[0] = root_result;

        dfs3(adj, 0, -1, result, n, count);

        return result;
    }
};
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