题解 | #在二叉树中找到两个节点的最近公共祖先#

#     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 o1 int整型 
# @param o2 int整型 
# @return int整型
#
class Solution:
    def __init__(self) -> None:
        self.paths = []
        self.pathhis = []

    def findp(self,root,p):
        if root is None:
            return False
        else:
            self.pathhis.append(root.val)
            if root.val==p:
                self.paths.append(self.pathhis)
                self.pathhis = []
                return True
            else:
                leftg = self.findp(root.left,p)
                rightg = self.findp(root.right,p)
                flag = leftg or rightg
                if flag is False:
                    self.pathhis.pop()
                return flag
    
    def lowestCommonAncestor(self , root: TreeNode, p: int, q: int) -> int:
        # write code here
        getp = self.findp(root,p)
        getq = self.findp(root,q)
        i = 0
        res = root.val
        # 比较两个路径，找到第一个不同的点
        path1 = self.paths[0]
        path2 = self.paths[1]
        while i < len(path1) and i < len(path2):
            if path1[i] == path2[i]:
                # 最后一个相同的节点就是最近公共祖先
                res = path1[i]
                i += 1
            else:
                break
        
        return res