# Time:  O(n)
# Space: O(n)

# An undirected, connected tree with N nodes
# labelled 0...N-1 and N-1 edges are given.
#
# The ith edge connects nodes edges[i][0] and edges[i][1] together.
#
# Return a list ans, where ans[i] is the sum of the distances
# between node i 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:
# Here is a diagram of the given tree:
#   0
#  / \
# 1   2
#    /|\
#   3 4 5
# 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.
# Note: 1 <= N <= 10000

import collections


class Solution(object):
    def sumOfDistancesInTree(self, N, edges):
        """
        :type N: int
        :type edges: List[List[int]]
        :rtype: List[int]
        """
        def dfs(graph, node, parent, count, result):
            for nei in graph[node]:
                if nei != parent:
                    dfs(graph, nei, node, count, result)
                    count[node] += count[nei]
                    result[node] += result[nei] + count[nei]

        def dfs2(graph, node, parent, count, result):
            for nei in graph[node]:
                if nei != parent:
                    result[nei] = result[node] - count[nei] + \
                                  len(count) - count[nei]
                    dfs2(graph, nei, node, count, result)

        graph = collections.defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)

        count = [1] * N
        result = [0] * N

        dfs(graph, 0, None, count, result)
        dfs2(graph, 0, None, count, result)
        return result

Create sum-of-distances-in-tree.py