tech-cow

class Solution(object):
    def diameterOfBinaryTree(self, root):
        if not root: return 0
        left = self.getHeight(root.left)
        right= self.getHeight(root.right)
        return max(left + right, max(self.diameterOfBinaryTree(root.left), self.diameterOfBinaryTree(root.right)))

    def getHeight(self, root):
        if not root: return 0
        left = self.getHeight(root.left)

tech-cow

Referral: Tiana Jihyung

tech-cow

正在面

潜心刷题....

电面挂

• Facebook
• Linkedin
• Airbnb (一共两面，挂在第二面)
• StackRox
• Coinbase (直接让你做TripleBytes选择题)

tech-cow

The Maze

定义:

determine whether the ball could stop at the destination.

这里对于stop的判定条件是：球从一个方向朝着end卷过来，能否在end点的下一个位置碰到墙。举个不成立的例子

比如你从左边冲过来，到达了end，虽然end点的上下两头都有墙，但右边没有墙，这种就不符合定义。

tech-cow

.

tech-cow

class Solution:
    def subsets(self, nums):
        if not nums: return []
        self.res = []
        self.dfs(nums, [], 0)
        return self.res
    
    
    def dfs(self, nums, temp, index):
        if index == len(nums):

tech-cow

def getHeight(root):
  if not root: return 0

  left = getHeight(root.left)
  right = getHeight(root.right)
  return max(left, right) + 1

tech-cow

# n1 -> n2 -> n3

Class Node(self):
	def __init__(self, val):
	      self.next = next
	      self.val = val

	def reverse_list(self, head):
	      prev = None
	      cur = head

tech-cow

def find_nearest(arr, target):
	l , r = 0, len(arr) - 1
	while l + 1 < r:
	      if arr[mid] == target:
		return mid
	      elif arr[mid] < target:
		l = mid
	      else:
		r = mid
	#preprocess

tech-cow

def find_first_occ_gd(arr , target):
l , r = 0, len(arr) - 1
while l + 1 < r:
	mid = l + (r - l) // 2
if arr[mid] == target:
r = mid
elif  arr[mid] < target:
l = mid + 1          #可以不加1，不影响当arr[mid] == target的时候 r = mid的设置
else: 
r = mid - 1           #可以不减1，不影响当arr[mid] == target的时候 r = mid的设置