LeetCode in Kotlin

1361. Validate Binary Tree Nodes

Medium

You have n binary tree nodes numbered from 0 to n - 1 where node i has two children leftChild[i] and rightChild[i], return true if and only if all the given nodes form exactly one valid binary tree.

If node i has no left child then leftChild[i] will equal -1, similarly for the right child.

Note that the nodes have no values and that we only use the node numbers in this problem.

Example 1:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,-1,-1,-1]

Output: true

Example 2:

Input: n = 4, leftChild = [1,-1,3,-1], rightChild = [2,3,-1,-1]

Output: false

Example 3:

Input: n = 2, leftChild = [1,0], rightChild = [-1,-1]

Output: false

Constraints:

• n == leftChild.length == rightChild.length
• 1 <= n <= 104
• -1 <= leftChild[i], rightChild[i] <= n - 1

Solution

import java.util.ArrayDeque
import java.util.Deque

class Solution {
    fun validateBinaryTreeNodes(n: Int, leftChild: IntArray, rightChild: IntArray): Boolean {
        val inDeg = IntArray(n)
        for (i in 0 until n) {
            if (leftChild[i] >= 0) {
                inDeg[leftChild[i]] += 1
            }
            if (rightChild[i] >= 0) {
                inDeg[rightChild[i]] += 1
            }
        }
        val queue: Deque<Int> = ArrayDeque()
        for (i in 0 until n) {
            if (inDeg[i] == 0) {
                if (queue.isEmpty()) {
                    queue.offer(i)
                } else {
                    // Violate rule 1.
                    return false
                }
            }
            if (inDeg[i] > 1) {
                // Violate rule 2.
                return false
            }
        }
        var tpLen = 0
        while (queue.isNotEmpty()) {
            val curNode = queue.poll()
            tpLen++
            val left = leftChild[curNode]
            val right = rightChild[curNode]
            if (left > -1 && --inDeg[left] == 0) {
                queue.offer(left)
            }
            if (right > -1 && --inDeg[right] == 0) {
                queue.offer(right)
            }
        }
        return tpLen == n
    }
}

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