LeetCode in Kotlin

834. Sum of Distances in Tree

Hard

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

class Solution {
    private var n = 0
    private lateinit var count: IntArray
    private lateinit var answer: IntArray
    private lateinit var graph: Array<MutableList<Int>?>

    private fun postorder(node: Int, parent: Int) {
        for (child in graph[node]!!) {
            if (child != parent) {
                postorder(child, node)
                count[node] += count[child]
                answer[node] += answer[child] + count[child]
            }
        }
    }

    private fun preorder(node: Int, parent: Int) {
        for (child in graph[node]!!) {
            if (child != parent) {
                answer[child] = answer[node] - count[child] + n - count[child]
                preorder(child, node)
            }
        }
    }

    fun sumOfDistancesInTree(n: Int, edges: Array<IntArray>): IntArray {
        this.n = n
        count = IntArray(n)
        answer = IntArray(n)
        graph = arrayOfNulls(n)
        count.fill(1)
        for (i in 0 until n) {
            graph[i] = ArrayList()
        }
        for (edge in edges) {
            graph[edge[0]]?.add(edge[1])
            graph[edge[1]]?.add(edge[0])
        }
        postorder(0, -1)
        preorder(0, -1)
        return answer
    }
}

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