题解 | #求二叉树的层序遍历#

/**
 * struct TreeNode {
 *  int val;
 *  struct TreeNode *left;
 *  struct TreeNode *right;
 * };
 */
/**
 * 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
 * @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
 */
int** levelOrder(struct TreeNode* root, int* returnSize,
                 int** returnColumnSizes) {
    *returnSize = 0;
    if (!root)
        return NULL;
    int** res = (int**)malloc(sizeof(int*) * 1500);
    struct TreeNode* que[1500];
    int head = 0;
    int rear = 0;
    que[rear++] = root;
    *returnColumnSizes = (int*)malloc(sizeof(int) * 1500);
    while (head != rear) {
        struct TreeNode* tmp;
        int prerear = rear;
        res[*returnSize] = (int*)malloc(sizeof(int) * (rear - head));
        int k = 0;
        while (head != prerear) {
            tmp = que[head++];
            res[(*returnSize)][k++] = tmp->val;
            if (tmp->left)
                que[rear++] = tmp->left;
            if (tmp->right)
                que[rear++] = tmp->right;
        }
        (*returnColumnSizes)[*returnSize] = k;
        (*returnSize)++;
    }
    return res;
}