834. Sum of Distances in Tree

Description

There is an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges.

You are given the integer n and the array edges where edges[i] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i in the tree.

Return an array answer of length n where answer[i] is the sum of the distances between the i^th node in the tree and all other nodes.

Example 1:

Input: n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
Output: [8,12,6,10,10,10]
Explanation: 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.

Example 2:

Input: n = 1, edges = []
Output: [0]

Example 3:

Input: n = 2, edges = [[1,0]]
Output: [1,1]

Constraints:

1 <= n <= 3 * 10⁴
edges.length == n - 1
edges[i].length == 2
0 <= a_i, b_i < n
a_i != b_i
The given input represents a valid tree.

Solution

sum-of-distances-in-tree.py

class Solution:
    def sumOfDistancesInTree(self, N, edges):
        tree = collections.defaultdict(set)
        res = [0] * N
        count = [1] * N
        for i, j in edges:
            tree[i].add(j)
            tree[j].add(i)

        def dfs(root, pre):
            for i in tree[root]:
                if i != pre:
                    dfs(i, root)
                    count[root] += count[i]
                    res[root] += res[i] + count[i]

        def dfs2(root, pre):
            for i in tree[root]:
                if i != pre:
                    res[i] = res[root] - count[i] + N - count[i]
                    dfs2(i, root)
        dfs(0, -1)
        dfs2(0, -1)
        return res