祖父节点值为偶数的节点和

这个解决方案使用深度优先搜索（DFS）来遍历整棵树，并在遍历过程中寻找符合条件的节点。首先检查当前节点值是否为偶数，如果是，那么它的孙子节点（如果存在）的值将被加到结果中。之后，递归地对左右子树应用同样的逻辑。这种方法确保了每个节点都被访问一次，且只有当节点的祖父节点的值为偶数时，节点的值才会被加入总和。

时间复杂度: O(n)

空间复杂度: O(n) in the worst case, O(log(n)) in the best case

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def sumEvenGrandparent(self, root: TreeNode) -> int:
        re = 0
        # Base case: if the node is a leaf, return 0
        if not root.left and not root.right:
            return 0
        # Check if the current node's value is even
        if root.val % 2 == 0:
            # Check left child's children
            if root.left:
                if root.left.left:
                    re += root.left.left.val
                if root.left.right:
                    re += root.left.right.val
            # Check right child's children
            if root.right:
                if root.right.left:
                    re += root.right.left.val
                if root.right.right:
                    re += root.right.right.val
        # Recursively sum up values for left subtree
        if root.left:
            re += self.sumEvenGrandparent(root.left)
        # Recursively sum up values for right subtree
        if root.right:
            re += self.sumEvenGrandparent(root.right)
        return re