# 找兩個節點的最近的公共祖先
# 假設兩個節點 node1 和 node2,那麼最近公共祖先和node1、node2存在以下關係：
# 1、node1,node2分別在祖先左右兩側
# 2、祖先是node1,node2在祖先左/右側
# 3、祖先是node2,node1在祖先左/右側
# 使用dfs深度優先遍歷進行求解
# 定義一顆二叉樹


class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None
# 根據list建立二叉樹


def listcreattree(root, llist, i):
    if i < len(llist):
        if llist[i] == '#':
            return None
        else:
            root = TreeNode(llist[i])
            root.left = listcreattree(root.left, llist, 2 * i + 1)
            root.right = listcreattree(root.right, llist, 2 * i + 2)
            return root
    return root


class Solution:
    def nearest_common_ancestor(self, root, node1, node2):
        if not root:
            return None
        if root.val == node1 or root.val == node2:
            return root.val
        left = self.nearest_common_ancestor(root.left, node1, node2)
        right = self.nearest_common_ancestor(root.right, node1, node2)
        if left is not None and right is not None:
            return root.val
        if left is None:
            return right
        if right is None:
            return left
        return None


if __name__ == '__main__':
    test_list = [3, 5, 1, 6, 2, 0, 8, '#', '#', 7, 4]
    root = listcreattree(None, test_list, 0)
    tmp1 = Solution()
    print(tmp1.nearest_common_ancestor(root, 5, 1))
# 結果為 3