遍历一次二叉树，查看是否有符合条件的两个元素。
暴力实现，不考虑剪枝，遍历顺序不重要。
class Solution:
def findTarget(self, root: Optional[TreeNode], k: int) -> bool:
if root is None:
return False
stack, visited = [root], set()
while stack:
node = stack.pop()
if k - node.val in visited:
return True
visited.add(node.val)
if node.right is not None:
stack.append(node.right)
if node.left is not None:
stack.append(node.left)
return False