思路:先遍历二叉树,为偶数的则向下找两个层加起来
class Solution {
public:
int an = 0;
int sumEvenGrandparent(TreeNode* root) {
bianli(root);
return an;
}
void findSon(int level, TreeNode *node){
if(level == 2){
an += node->val;
// cout << an << endl;
return;
}else{
if(node->left != nullptr){
findSon(level+1, node->left);
}
if(node->right != nullptr){
findSon(level+1, node->right);
}
}
}
void bianli(TreeNode *node){
if(node->val % 2 == 0){
findSon(0, node);
}
if(node->left != nullptr){
bianli(node->left);
}
if(node->right != nullptr){
bianli(node->right);
}
}
};



