2385.js

class TreeNode {
  constructor(val = 0, left = null, right = null) {
    this.val = val
    this.left = left
    this.right = right
  }
}

const buildTree = (array, index = 0) => {
  if (index < array.length && array[index] != null) {
    let node = new TreeNode(arr[index])
    node.left = buildTree(array, 2 * index + 1)
    node.right = buildTree(array, 2 * index + 2)

    return node
  }

  return null
}

/** ------------------------------------------------------------------------------
 * 
 * 2385. Amount of Time for Binary Tree to Be Infected
 * Topics: Hash Table, DFS
 * https://leetcode.com/problems/amount-of-time-for-binary-tree-to-be-infected/description/
 * 
 ------------------------------------------------------------------------------ */
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} start
 * @return {number}
 */
var amountOfTime = function (root, start) {
  const graph = createGraph(root, {})

  console.log(graph)
}

let rootArray = [1, 5, 3, null, 4, 10, 6, 9, 2]
let start = 3

let root = buildTree(rootArray)
console.log(root, start)

const createGraph = (node, graph) => {
  if (!node) return graph

  if (!(node.val in graph)) graph[node.val] = []

  if (node.left) {
    graph[node.val].push(node.left.val)
    if (!node.left.val in graph) graph[node.left.val] = []
    graph[node.left.val].push(node.val)
    createGraph(node.left, graph)
  }

  if (node.right) {
    graph[node.val].push(node.right.val)
    if (!(node.right.val in graph)) graph[node.right.val] = []
    graph[node.right.val].push(node.val)
    createGraph(node.right, graph)
  }
  return graph
}