文件: root - text - article - 2019 - 10 - floor-and-ceiling-of-a-binary-search-tree.txt
标签: 每日算法题, 算法, 数据结构, 面试题, Daily Problem, Data Structures and Algorithms, | 英文 | 主页 | 类别: 计算机科学 | 373 次阅读, 24840 次搜索 | 119 个单词
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标签: 每日算法题, 算法, 数据结构, 面试题, Daily Problem, Data Structures and Algorithms, | 英文 | 主页 | 类别: 计算机科学 | 373 次阅读, 24840 次搜索 | 119 个单词
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Hi, here's your problem today. This problem was recently asked by Apple:
Given an integer k and a binary search tree, find the floor (less than or equal to) of k, and the ceiling (larger than or equal to) of k. If either does not exist, then print them as None.
Here is the definition of a node for the tree.
标签: 每日算法题, 算法, 数据结构, 面试题, Daily Problem, Data Structures and Algorithms, | 英文 | 主页 | 类别: 计算机科学 | 373 次阅读, 24840 次搜索 | 119 个单词 Given an integer k and a binary search tree, find the floor (less than or equal to) of k, and the ceiling (larger than or equal to) of k. If either does not exist, then print them as None.
Here is the definition of a node for the tree.
class Node:
def __init__(self, value):
self.left = None
self.right = None
self.value = value
def findCeilingFloor(root_node, k, floor=None, ceil=None):
# Fill this in.
root = Node(8)
root.left = Node(4)
root.right = Node(12)
root.left.left = Node(2)
root.left.right = Node(6)
root.right.left = Node(10)
root.right.right = Node(14)
print findCeilingFloor(root, 5)
# (4, 6)
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