题解 | #二叉搜索树的最近公共祖先#
二叉搜索树的最近公共祖先
http://www.nowcoder.com/practice/d9820119321945f588ed6a26f0a6991f
#coding:utf-8
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
#
# @param root TreeNode类
# @param p int整型
# @param q int整型
# @return int整型
#
class Solution:
def lowestCommonAncestor(self , root , p , q ):
# write code here
def lowestRec(root,o1,o2):
if root is None or root.val == o1 or root.val == o2:
return root
if p < root.val and q < root.val:
return lowestRec(root.left, o1, o2)
elif (p > root.val and q > root.val):
return lowestRec(root.right, o1, o2)
else:
return root
return root
return lowestRec(root, p, q).val
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
#
# @param root TreeNode类
# @param p int整型
# @param q int整型
# @return int整型
#
class Solution:
def lowestCommonAncestor(self , root , p , q ):
# write code here
def lowestRec(root,o1,o2):
if root is None or root.val == o1 or root.val == o2:
return root
if p < root.val and q < root.val:
return lowestRec(root.left, o1, o2)
elif (p > root.val and q > root.val):
return lowestRec(root.right, o1, o2)
else:
return root
return root
return lowestRec(root, p, q).val