题解 | #二叉树的前序遍历#
二叉树的前序遍历
https://www.nowcoder.com/practice/5e2135f4d2b14eb8a5b06fab4c938635
/**
* 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
*/
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
static int num;
void preoder(struct TreeNode* root, int* a)
{
if(root == NULL)
return;
a[num++] = root->val;
preoder(root->left,a);
preoder(root->right,a);
}
int* preorderTraversal(struct TreeNode* root, int* returnSize ) {
//二叉树的节点数量满足 1≤n≤100
int *a = (int *)malloc(sizeof(int)*100);
preoder(root,a);
*returnSize = num;
return a;
}
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