题解 | #牛群的二叉树排序#

/**
 * 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 cows int整型一维数组
 * @param cowsLen int cows数组长度
 * @return TreeNode类
 */
struct TreeNode* createTree(int type, int count) {
    struct TreeNode* node[count];
    if(count == 0) return NULL;
    for (int i = 0; i < count; i++) {
        node[i] = malloc(sizeof(struct TreeNode));
        node[i]->val = type;
        node[i]->left = NULL;
        node[i]->right = NULL;
    }
    for (int i = 0; i < count; i++) {
        if (2 * i + 1 < count) {
            node[i]->left = node[2 * i + 1];
        } else node[i]->left = NULL;
        if (2 * i + 2 < count) {
            node[i]->right = node[2 * i + 2];
        } else node[i]->right = NULL;
    }
    return node[0];
}
struct TreeNode* sortCowsTree(int* cows, int cowsLen ) {
    int count = 0;
    for (int i = 0; i < cowsLen; i++) {
        if (cows[i] == 0) count++;
    }
    struct TreeNode* root = malloc(sizeof(struct TreeNode));
    root->val = -1;
    root->left = createTree(0, count);
    root->right = createTree(1, cowsLen - count);
    return root;
}