Validate Binary Tree Nodes

Description

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

Example 4:

Input: n = 6, leftChild = [1,-1,-1,4,-1,-1], rightChild = [2,-1,-1,5,-1,-1]
Output: false

 

Constraints:

• 1 <= n <= 10^4
• leftChild.length == rightChild.length == n
• -1 <= leftChild[i], rightChild[i] <= n - 1

Solution(javascript)

/* eslint-disable no-loop-func */
/**
 * @param {number} n
 * @param {number[]} leftChild
 * @param {number[]} rightChild
 * @return {boolean}
 */
const validateBinaryTreeNodes = function (n, leftChild, rightChild) {
  const used = new Array(n).fill(false)
  let valid = true
  const bfs = (index) => {
    let current = [index]
    used[index] = true
    while (current.length > 0) {
      const next = []
      current.forEach((currentIndex) => {
        if (used[leftChild[currentIndex]] || used[rightChild[currentIndex]]) {
          valid = false
          return
        }
        if (leftChild[currentIndex] !== -1) {
          used[leftChild[currentIndex]] = true
          next.push(leftChild[currentIndex])
        }
        if (rightChild[currentIndex] !== -1) {
          used[rightChild[currentIndex]] = true
          next.push(rightChild[currentIndex])
        }
      })
      current = next
    }
  }
  bfs(0)
  return valid && used.every(x => x)
}