LeetcodeT1315.祖父結點爲偶數的結點和

代碼：

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

/*
    算法概述：先序遍歷二叉樹，若當前結點爲偶數，則分別累加它的左右孩子結點的左右孩子結點值（如果有的話）
*/
int sumEvenGrandparent(struct TreeNode* root){
    int sum = 0;

    if (root == NULL)// 空二叉樹
        return 0;
    if (root -> left == NULL && root -> right == NULL)// 只有一層的二叉樹
        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)))// 有兩層的二叉樹
        return 0;
    
    if ((root -> val) % 2 == 0) {
        if (root -> left != NULL) {// 統計左子樹上滿足條件的
            if (root -> left -> left != NULL)
                sum += root -> left -> left -> val;
            if (root -> left -> right != NULL)
                sum += root -> left -> right -> val;    
        }
        if (root -> right != NULL) {// 統計右子樹上滿足條件的
            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;
}

兩層二叉樹那裏看着有點複雜……