题解 | #二叉树# #二叉搜索树的最近公共祖先#

#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
        if root == None or p == None or q == None:
            return root
        #switch node
        if p < q:
            tmp = p
            p = q
            q = tmp

        ##coner case

        #cond1: root == p or root == q
        #cond2: root.val > p.val and root.val < q.val
        #cond3: root.val > p.val
        #cond4: root.val < p.val
        print ("root.val, p, q: ", root.val, p, q)
        if root.val == p:
            print ("ck1")
            return root.val
        elif root.val > q and root.val < p:
            print ("ck2")
            return root.val
        elif root.val > p:
            print ("ck3")
            return self.lowestCommonAncestor(root.left, p, q)
        else:
            print ("ck4")
            return self.lowestCommonAncestor(root.right, p, q)

        return root.val