231、2的幂
class Solution:
def isPowerOfTwo(self, n: int) -> bool:
return n > 0 and not (n & (n - 1))
235、二叉搜索树的最近公共祖先
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if p.val<root.val and q.val<root.val:
return self.lowestCommonAncestor(root.left,p,q)
if p.val>root.val and q.val>root.val:
return self.lowestCommonAncestor(root.right,p,q)
return root
236、二叉搜索树的最近公共祖先
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if root is None:
return root
if root is p or root is q:
return root
left=self.lowestCommonAncestor(root.left,p,q)
right=self.lowestCommonAncestor(root.right,p,q)
if left and right:
return root
if left:
return left
if right:
return right
