<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:
    vector<int> subtreeSum;
    vector<bool> vis;
    int dfs(unordered_map<int, vector<int>>& gr, int source) {
        vis[source] = true;
        int currSum = 1;
        for (int child : gr[source]) {
            if (!vis[child]) {
                currSum += dfs(gr, child);
            }
        }
        return subtreeSum[source] = currSum;
    }
    int bfsRoot(unordered_map<int, vector<int>>& gr) {
        unordered_map<int, int> level;
        level[0] = 0;
        vis[0] = true;
        queue<int> q;
        q.push(0);
        int distance = 0;
        while (!q.empty()) {
            int node = q.front();
            q.pop();
            for (int nbr : gr[node]) {
                if (!vis[nbr]) {
                    vis[nbr] = true;
                    q.push(nbr);
                    level[nbr] = level[node] + 1;
                    distance += level[nbr];
                }
            }
        }
        return distance;
    }
    void bfs(unordered_map<int, vector<int>>& gr, int source, vector<int>& ans,
             int n) {
        queue<int> q;
        q.push(source);
        vis[source] = true;
        while (!q.empty()) {
            int node = q.front();
            q.pop();
            for (int nbr : gr[node]) {
                if (!vis[nbr]) {
                    vis[nbr] = true;
                    q.push(nbr);
                    ans[nbr] =
                        ans[node] - subtreeSum[nbr] + (n - subtreeSum[nbr]);
                }
            }
        }
    }
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        if (n == 1) {
            vector<int> ans;
            ans.push_back(0);
            return ans;
        }
        unordered_map<int, vector<int>> gr;
        for (auto v : edges) {
            gr[v[0]].push_back(v[1]);
            gr[v[1]].push_back(v[0]);
        }
        subtreeSum.resize(n, 0);
        vis.resize(n, false);
        dfs(gr, 0);
        for (int i = 0; i < n; i++) {
            vis[i] = false;
        }
        int rootAns = bfsRoot(gr);
        vector<int> ans(n);
        ans[0] = rootAns;
        for (int i = 0; i < n; i++) {
            vis[i] = false;
        }
        bfs(gr, 0, ans, n);
        return ans;
    }
};
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