<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:
    vector<int> graph[200001];
    ll dp[200001], ans[200001];

    void dfs1(int node, int parent, ll depth) {
        ans[1] += depth;
        dp[node] = 1;
        for (int i : graph[node]) {
            if (i != parent) {
                dfs1(i, node, depth + 1);
                dp[node] += dp[i];
            }
        }
    }

    void dfs2(int node, int parent, int n) {
        for (int i : graph[node]) {
            if (i != parent) {
                ans[i] = ans[node] + n - 2 * dp[i];
                dfs2(i, node, n);
            }
        }
    }

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        
        for (int i = 1; i <= n; i++) {
            graph[i].clear();
            dp[i] = ans[i] = 0;
        }

        
        for (auto& edge : edges) {
            graph[edge[0] + 1].push_back(edge[1] + 1);
            graph[edge[1] + 1].push_back(edge[0] + 1);
        }

       
        dfs1(1, 0, 0); 
        dfs2(1, 0, n); 

        
        vector<int> result;
        for (int i = 1; i <= n; i++) result.push_back(ans[i]);
        return result;
    }
};

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