Medium
There is a binary tree rooted at 0
consisting of n
nodes. The nodes are labeled from 0
to n - 1
. You are given a 0-indexed integer array parents
representing the tree, where parents[i]
is the parent of node i
. Since node 0
is the root, parents[0] == -1
.
Each node has a score. To find the score of a node, consider if the node and the edges connected to it were removed. The tree would become one or more non-empty subtrees. The size of a subtree is the number of the nodes in it. The score of the node is the product of the sizes of all those subtrees.
Return the number of nodes that have the highest score.
Example 1:
Input: parents = [-1,2,0,2,0]
Output: 3
Explanation:
The score of node 0 is: 3 * 1 = 3
The score of node 1 is: 4 = 4
The score of node 2 is: 1 * 1 * 2 = 2
The score of node 3 is: 4 = 4
The score of node 4 is: 4 = 4
The highest score is 4, and three nodes (node 1, node 3, and node 4) have the highest score.
Example 2:
Input: parents = [-1,2,0]
Output: 2
Explanation:
The score of node 0 is: 2 = 2
The score of node 1 is: 2 = 2
The score of node 2 is: 1 * 1 = 1
The highest score is 2, and two nodes (node 0 and node 1) have the highest score.
Constraints:
n == parents.length
2 <= n <= 105
parents[0] == -1
0 <= parents[i] <= n - 1
for i != 0
parents
represents a valid binary tree.class Solution {
internal class Node {
var left: Node? = null
var right: Node? = null
}
private var size = 0
private var max: Long = 0
private var freq = 0
private fun postOrder(root: Node?): Long {
if (root == null) {
return 0
}
val left = postOrder(root.left)
val right = postOrder(root.right)
val `val` = Math.max(1, left) * Math.max(1, right) * Math.max(size - left - right - 1, 1)
if (`val` > max) {
max = `val`
freq = 1
} else if (`val` == max) {
freq += 1
}
return left + right + 1
}
fun countHighestScoreNodes(parents: IntArray): Int {
size = parents.size
val nodes = arrayOfNulls<Node>(size)
for (i in 0 until size) {
nodes[i] = Node()
}
var root: Node? = null
for (i in 0 until size) {
if (parents[i] != -1) {
val node = nodes[parents[i]]
if (node!!.left == null) {
node.left = nodes[i]
} else {
node.right = nodes[i]
}
} else {
root = nodes[i]
}
}
postOrder(root)
return freq
}
}