LeetcodeT1315.祖父结点为偶数的结点和

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
*/

/*
    算法概述：先序遍历二叉树，若当前结点为偶数，则分别累加它的左右孩子结点的左右孩子结点值（如果有的话）
*/
int sumEvenGrandparent(struct TreeNode* root){
    int sum = 0;

    if (root == NULL)// 空二叉树
        return 0;
    if (root -> left == NULL && root -> right == NULL)// 只有一层的二叉树
        return 0;
    if (((root -> left != NULL && root -> left -> left == NULL && root -> left -> right == NULL) && (root -> right == NULL)) || 
        ((root -> right != NULL && root -> right -> left == NULL && root -> right -> right == NULL) && (root -> left == NULL)) ||   
        ((root -> left != NULL && root -> left -> left == NULL && root -> left -> right == NULL) && 
         (root -> right != NULL && root -> right -> left == NULL && root -> right -> right == NULL)))// 有两层的二叉树
        return 0;
    
    if ((root -> val) % 2 == 0) {
        if (root -> left != NULL) {// 统计左子树上满足条件的
            if (root -> left -> left != NULL)
                sum += root -> left -> left -> val;
            if (root -> left -> right != NULL)
                sum += root -> left -> right -> val;    
        }
        if (root -> right != NULL) {// 统计右子树上满足条件的
            if (root -> right -> left != NULL)
                sum += root -> right -> left -> val;
            if (root -> right -> right != NULL)
                sum += root -> right -> right -> val;    
        }
    }
    sum += sumEvenGrandparent(root -> left) + sumEvenGrandparent(root -> right);

    return sum;
}

两层二叉树那里看着有点复杂……