<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 to store the count of nodes in the subtree rooted at each node
    vector<int> count;
    // Vector to store the result, i.e., sum of distances from each node to all other nodes
    vector<int> res;
    // Adjacency list to store the tree
    unordered_map<int, vector<int>> adj;

    // First DFS to calculate the count of nodes and initial distances
    void dfs(int cur, int par) {
        // Traverse all the children of the current node
        for (int child : adj[cur]) {
            if (child != par) {  // Avoid revisiting the parent node
                dfs(child, cur);  // Recursive DFS call for the child node
                count[cur] += count[child];  // Update the count of nodes in the current subtree
                res[cur] += res[child] + count[child];  // Update the result for the current node
            }
        }
    }

    // Second DFS to compute the result for all nodes based on the root result
    void dfs2(int cur, int par) {
        // Traverse all the children of the current node
        for (int child : adj[cur]) {
            if (child != par) {  // Avoid revisiting the parent node
                // Compute the result for the child node based on the current node's result
                res[child] = res[cur] - count[child] + (count.size() - count[child]);
                dfs2(child, cur);  // Recursive DFS call for the child node
            }
        }
    }

    // Main function to calculate the sum of distances in the tree
    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        // Clear adjacency list for the new input
        adj.clear();
        // Initialize count of nodes in the subtree for each node to 1 (counting the node itself)
        count = vector<int>(n, 1);
        // Initialize the result vector with 0s
        res = vector<int>(n, 0);

        // Construct the adjacency list from the given edges
        for (auto e : edges) {
            adj[e[0]].push_back(e[1]);
            adj[e[1]].push_back(e[0]);
        }

        // Run the first DFS from node 0 (root) with -1 as its parent (indicating no parent)
        dfs(0, -1);

        // Run the second DFS to compute the results for all nodes
        dfs2(0, -1);

        return res;  // Return the result vector
    }
};

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