863. All Nodes Distance K in Binary Tree

var distanceK = function (root, target, K) {
	root.parent = null

	function findKNodes(node, K) {
		const queue = [node],
			result = []
		while (queue.length) {
			const curr = queue.shift(),
				level = curr.level,
				left = curr.left,
				right = curr.right,
				parent = curr.parent

			if (level === K) {
				result.push(curr.val)
			} else {
				if (left && left.level === undefined) {
					left.level = level + 1
					queue.push(left)
				}
				if (right && right.level === undefined) {
					right.level = level + 1
					queue.push(right)
				}
				if (parent && parent.level === undefined) {
					parent.level = level + 1
					queue.push(parent)
				}
			}
		}
		return result
	}

	function bfsSearch(node, K) {
		const queue = [node]
		while (queue.length) {
			const curr = queue.shift()
			if (curr.val === target.val) {
				curr.level = 0
				return findKNodes(curr, K)
			}
			const left = curr.left,
				right = curr.right
			if (left) {
				left.parent = curr
				queue.push(left)
			}
			if (right) {
				right.parent = curr
				queue.push(right)
			}
		}
	}
	return bfsSearch(root, K)
}

const root = {
	val: 3,
	right: {
		val: 1,
		right: { val: 8, right: null, left: null },
		left: { val: 0, right: null, left: null },
	},
	left: {
		val: 5,
		right: {
			val: 2,
			right: { val: 4, right: null, left: null },
			left: { val: 7, right: null, left: null },
		},
		left: { val: 6, right: null, left: null },
	},
}

//Test cases
console.log(
	distanceK(
		root,
		{
			val: 5,
			right: {
				val: 2,
				right: { val: 4, right: null, left: null },
				left: { val: 7, right: null, left: null },
			},
			left: { val: 6, right: null, left: null },
		},
		2
	)
)