题解 | #二叉搜索树的最近公共祖先#

# 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 __init__(self) -> None:
        self.minc = 0
        self.minanc = 0

    def findp(self,root,p):
        if root is None:
            return False
        else:
            if root.val==p:
                return True
            else:
                leftg = self.findp(root.left,p)
                rightg = self.findp(root.right,p)
                return leftg or rightg

    def dps(self, root, p,q, depth):
        if root is None:
                return 0
        else:
            getp = self.findp(root,p)
            getq = self.findp(root,q)
            if getp and getq:
                if depth>self.minc:
                    self.minc = depth
                    self.minanc = root.val
            leftg = self.dps(root.left,p,q,depth+1)
            rightg = self.dps(root.right,p,q,depth+1)   
            return 0

            
    def lowestCommonAncestor(self , root: TreeNode, p: int, q: int) -> int:
        # write code here
        self.dps(root,p,q,1)
        return self.minanc