题解 | #二叉树中和为某一值的路径(二)#

/**
 * struct TreeNode {
 *    int val;
 *    struct TreeNode *left;
 *    struct TreeNode *right;
 * };
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param root TreeNode类 
 * @param target int整型 
 * @return int整型二维数组
 * @return int* returnSize 返回数组行数
 * @return int** returnColumnSizes 返回数组列数
 */
static int a[5000];
static int i = 0;
void inorder_calc(struct TreeNode *root, int target, int* returnSize, int** returnColumnSizes, int **array)
{
    int sum = 0;
    if (root == NULL) {
        return;
    }
    a[i++] = root->val;
    if (root->left == NULL && root->right == NULL) {
        for (int m = 0; m < i; m++) {
            sum += a[m];
        }
        if(sum == target) {
            array[(*returnSize)] = (int *) malloc(sizeof(int) * i);
            memcpy(array[(*returnSize)], a, i * sizeof(int));
            (*returnColumnSizes)[(*returnSize)] = i;
            (*returnSize)++;
        }
        i--;
        return;
    }
    inorder_calc(root->left, target, returnSize, returnColumnSizes, array);
    inorder_calc(root->right, target, returnSize, returnColumnSizes, array);
    i--;
}

int** FindPath(struct TreeNode* root, int target, int* returnSize, int** returnColumnSizes ) {
    // write code here
    *returnSize = 0;
    int **array = (int **)malloc(sizeof(int*) * 5000);
    (*returnColumnSizes) = (int *)malloc(sizeof(int) * 5000);
    inorder_calc(root, target, returnSize, returnColumnSizes, array);
//     printf("rc=%d\n", (*returnColumnSizes)[0]);
//     printf("%d,%d\n",array[0][0], array[0][1]);
    return array;
}