Leetcode: 二叉树的层次遍历

二叉树的层次遍历

---

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

这个题的难点并不在于层次遍历，层次遍历一个队列就足够了，难道在于需要按层次进行存储，我一开始的时候，一直用了一个二维数组，结果内存超过限制，因为事实上每一次的节点数很不同，所以不能直接设置固定的。
C语言代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *columnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
#define SIZE 999
int** levelOrder(struct TreeNode* root, int** columnSizes, int* returnSize) {
    int **result = (int **)malloc(sizeof(int* )*SIZE);
    struct TreeNode *queue[SIZE]; //定于队列
    int rear = 0;
    int head = 0; // 首位开始都在0，此时为空
    int *key =  (int*)malloc(sizeof(int)*SIZE);
    memset(key,0,sizeof(int)*SIZE);
    if(root == NULL)
        return NULL;
    queue[rear] = root;
    rear = (rear + 1) % SIZE;
    int height = 0;
    key[height] += 1;
    while(head != rear){
        height += 1;
        int i = 0;
        printf("malloc: %d ",key[height-1]);
        result[height-1] = (int *)malloc(sizeof(int) * key[height-1]);
        printf("%d %d
",head,rear);
        while(i < key[height-1]){
            printf("debug");
            root = queue[head];
            head =(head + 1) % SIZE;
            result[height-1][i++] = root->val;
            if(root->left){
                queue[rear] = root->left;
                rear = (rear + 1) % SIZE;
                key[height] += 1;
            }
            if(root->right){
                queue[rear] = root->right;
                rear = (rear + 1) % SIZE;
                printf("height: %d", height);
                key[height] += 1;
            }
        }
    }
    printf("last");
    *columnSizes = key;
    *returnSize = height;
    return result;
}