<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 result = 0;
    int N;
    vector<int> count;

    int dfsbase(vector<int> adj[], int curr, int prev, int currdepth) {
        int counting = 1; // The number of nodes in the subtree rooted at `curr`

        result += currdepth; // Add the current depth to the result for node 0
        
        // Traverse the adjacent nodes (children)
        for (int &child : adj[curr]) {
            if (child == prev) continue; // Skip the parent node
            // Recursively calculate for children
            counting += dfsbase(adj, child, curr, currdepth + 1);
        }

        count[curr] = counting; // Store the size of the subtree rooted at `curr`
        return counting; // Return the total size of the subtree
    }

    void dfs(vector<int> adj[], vector<int>& mainres, int parent, int prev) {
        // Traverse all children of the current parent
        for (int &child : adj[parent]) {
            if (child == prev) continue; // Skip the parent node

            // Calculate the result for the child based on the parent
            mainres[child] = mainres[parent] - count[child] + (N - count[child]);

            // Recursively call DFS for the child
            dfs(adj, mainres, child, parent);
        }
    }

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        N = n;
        count.resize(n, 0);
        vector<int> adj[n];

        // Build adjacency list
        for (vector<int>& vec : edges) {
            int u = vec[0];
            int v = vec[1];
            adj[u].push_back(v);
            adj[v].push_back(u);
        }

        // First DFS to calculate the base result for node 0
        dfsbase(adj, 0, -1, 0);

        vector<int> mainres(n, 0);
        mainres[0] = result; // Set the base result for node 0

        // Second DFS to calculate results for all other nodes
        dfs(adj, mainres, 0, -1);

        return mainres;
    }
};

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