题解 | 二叉树的中序遍历

/**
 * 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整型一维数组
 * @return int* returnSize 返回数组行数
 */
//中序遍历
void printnode(struct TreeLinkNode* node, int* num, int* returnSize) {
    if (node == NULL) {
        return;
    }
    printnode(node->left, num, returnSize);
    num[(*returnSize)++] = node->val;
    printnode(node->right, num, returnSize);
}

//为了避免越界，我的思路是先计算节点，避免资源的浪费
void countNodes(struct TreeNode* node, int* count) {
    if (node == NULL) {
        return;
    }
    countNodes(node->left, count);
    (*count)++;
    countNodes(node->right, count);
}

int* inorderTraversal(struct TreeNode* root, int* returnSize ) {
    // write code here
    countNodes(root, returnSize);
    int* ret = (int*)malloc(sizeof(int) * (*returnSize));
    *returnSize=0;
    printnode(root, ret, returnSize);
    return ret;
}