<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:
    def sumOfDistancesInTree(self, n: int, edges: List[List[int]]) -> List[int]:
        # similar to Time Taken to Mark All Nodes - rerooting
        import collections
        graph = collections.defaultdict(list)
        for a,b in edges:
            graph[a].append(b), graph[b].append(a)
        vals, nums, ans = [0] * n, [0] * n, [0] * n
        # nums[i]: no of nodes starting from root i, including i
        # vals[i]: sum of dist between i and all children (excluding parent.)
        # key part: find the parent nums and vals.

        self.dfs(0, None, graph, vals, nums)
        #print(vals, nchd)
        self.reroot(0, None, 0, 0, graph, vals, nums, ans)
        return ans

    def dfs(self, x, parent, graph, vals, nums):
        sum_v, sum_c = 0, 1
        for y in graph[x]:
            if y==parent: continue
            v, c = self.dfs(y,x,graph, vals, nums)
            sum_c += c
            sum_v += v
        vals[x] = sum_v + sum_c-1
        nums[x] = sum_c
        return vals[x], nums[x]

    def reroot(self, x, parent, pv, pc, graph, vals, nums, ans):
        
        ans[x] = pv + pc + vals[x]
        #print(x, parent, ':', ans[x], pv, pc)

        for y in graph[x]:
            if y==parent: continue
            # we can get others by simple computation
            others_chd = nums[x] - nums[y]
            others_val = vals[x] - vals[y] -nums[y]

            pv_ = pv + pc + others_val
            pc_ = pc + others_chd

            self.reroot(y, x, pv_, pc_, graph, vals, nums, ans)

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