<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<vector<int>> adj;
    vector<int> tree;
    vector<int> dist;
    vector<int> dp;
    vector<int> visited;
    vector<int> parent;

    // DFS to calculate tree size and initial distances
    void dfs(int start) {
        visited[start] = 1;
        tree[start] = 1; // Each node is at least its own subtree
        for(auto &child: adj[start]) {
            if(!visited[child]) {
                parent[child] = start;
                dist[child] = 1 + dist[start]; // Distance from root
                dfs(child);
                tree[start] += tree[child]; // Update parent’s subtree size
            }
        }
    }

    // Solve for dp values based on parent dp value
    void solve(int start, int n) {
        for(auto &child : adj[start]) {
            if (child != parent[start]) { // Prevent revisiting parent node
                dp[child] = dp[start] + n - 2 * tree[child];
                solve(child, n);
            }
        }
    }

    vector<int> sumOfDistancesInTree(int n, vector<vector<int>>& edges) {
        // Initialize all vectors
        adj.resize(n);
        dp.resize(n); // dp[i] : distance of i from all other vertices 
        tree.resize(n, 0); // tree[i] : number of nodes in subtree of i 
        dist.resize(n); // distance of i from root (0)
        dist[0] = 0;
        visited.resize(n, 0);
        parent.resize(n, -1); // Initialize parent as -1 for root

        // Build adjacency list
        for(auto &edge : edges) {
            adj[edge[0]].push_back(edge[1]);
            adj[edge[1]].push_back(edge[0]);
        }

        // Run DFS from root (0)
        dfs(0);

        // Compute initial dp[0] (sum of distances from node 0)
        for(int i = 0; i < n; i++) {
            dp[0] += dist[i];
        }

        // Compute dp values for other nodes
        solve(0, n);

        return dp;
    }
};

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