leetcode做题笔记103. 二叉树的锯齿形层序遍历

给你二叉树的根节点 root ，返回其节点值的 锯齿形层序遍历 。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。

思路一：BFS

#define N 2000

int** zigzagLevelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    if (root == NULL) {
        return NULL;
    }
    int** ans = malloc(sizeof(int*) * N);
    *returnColumnSizes = malloc(sizeof(int) * N);
    struct TreeNode* nodeQueue[N];
    int left = 0, right = 0;
    nodeQueue[right++] = root;
    bool isOrderLeft = true;

    while (left < right) {
        int levelList[N * 2];
        int front = N, rear = N;
        int size = right - left;
        for (int i = 0; i < size; ++i) {
            struct TreeNode* node = nodeQueue[left++];
            if (isOrderLeft) {
                levelList[rear++] = node->val;
            } else {
                levelList[--front] = node->val;
            }
            if (node->left) {
                nodeQueue[right++] = node->left;
            }
            if (node->right) {
                nodeQueue[right++] = node->right;
            }
        }
        int* tmp = malloc(sizeof(int) * (rear - front));
        for (int i = 0; i < rear - front; i++) {
            tmp[i] = levelList[i + front];
        }
        ans[*returnSize] = tmp;
        (*returnColumnSizes)[*returnSize] = rear - front;
        (*returnSize)++;
        isOrderLeft = !isOrderLeft;
    }
    return ans;
}

分析：

本题与上题相似，直接使用广度优先搜索将每层数放入数组再输出即可，注意        (*returnColumnSizes)[*returnSize] = rear - front;

总结：

本题考察广度优先搜索算法，将每层按左向右再右向左的顺序放入数组再输出即可