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推廣網站開發,包含 Laravel 和 Kotlin 後端撰寫、自動化測試、讀書心得等。Taiwan Kotlin User Group 管理員。

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Hi, here's your problem today. This problem was recently asked by Amazon:

Given a number n, generate all binary search trees that can be constructed with nodes 1 to n.

Here's some code to start with:

class Node:
  def __init__(self, value, left=None, right=None):
    self.value = value
    self.left = left
    self.right = right

  def __str__(self):
    result = str(self.value)
    if self.left:
      result = result + str(self.left)
    if self.right:
      result = result + str(self.right)
    return result

def generate_bst(n):
  # Fill this in.

for tree in generate_bst(3):
  print tree

# Pre-order traversals of binary trees from 1 to n.
# 123
# 132
# 213
# 312
# 321

#   1      1      2      3      3
#    \      \    / \    /      /
#     2      3  1   3  1      2
#      \    /           \    /
#       3  2             2  1