题解 | #从上往下打印二叉树#
从上往下打印二叉树
https://www.nowcoder.com/practice/7fe2212963db4790b57431d9ed259701
利用队列，层序遍历思想。利用结构数组创建队列，将当前节点的值存入数组；只要有左右孩子，就把孩子入队，当队列中头尾相遇时，意味着队列遍历结束，跳出。让returnsize=数组长度就好了
/**
* 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
*/
int* PrintFromTopToBottom(struct TreeNode* root, int* returnSize ) {
// write code here
int *arr = (int*)malloc(sizeof(int) * 1001);
struct TreeNode *Que[1001];
if(!root)return arr;
int front = 0, rear =0;
int len = 0;
Que[rear++] = root;
while(front != rear){
struct TreeNode *p = Que[front];
arr[len++] = p -> val;
if(p -> left){
Que[rear++] = p -> left;
}
if(p -> right){
Que[rear++] = p -> right;
}
front++;
}
*returnSize = len;
return arr;
}
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