Leetcode 100.相同的树

1.题目要求，如图所示:

我们可以用两个数组去解决此题:
1.首先我们要用malloc函数去构造两个数组，还要去构造两个数组的长度,代码块如下图所示:

	int* p_length = (int*)malloc(sizeof(int));
    *p_length = 0;
    int* q_length = (int*)malloc(sizeof(int));
    *q_length = 0;
    int* number_p = (int*)malloc(sizeof(int) * 1000);
    int* number_q = (int*)malloc(sizeof(int) * 1000);

2.我们再使用前序遍历(什么遍历都行)把树中的元素存入数组,代码块如下图所示:

void preorder_traver(struct TreeNode* root,int* number,int* length)
 {
    if(root == NULL)
    {
        number[(*length)++] = 10001;
        //因为存入数组的元素的相同，但路径可能不同，所以我们要存入一个超过范围数来判断它的路径
        return;
    }
    number[(*length)++] = root->val;
    preorder_traver(root->left,number,length);
    preorder_traver(root->right,number,length);
 }

3.再进行判断:

int i = 0;
    if((*p_length) != (*q_length))
        return false;
    for(i = 0;i < (*p_length);i++)
    {
        if(number_p[i] != number_q[i])
        {
            return false;
        }
    }
    return true;

全部代码如下图所示:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
 void preorder_traver(struct TreeNode* root,int* number,int* length)
 {
    if(root == NULL)
    {
        number[(*length)++] = 10001;
        return;
    }
    number[(*length)++] = root->val;
    preorder_traver(root->left,number,length);
    preorder_traver(root->right,number,length);
 }
bool isSameTree(struct TreeNode* p, struct TreeNode* q) {
    int* p_length = (int*)malloc(sizeof(int));
    *p_length = 0;
    int* q_length = (int*)malloc(sizeof(int));
    *q_length = 0;
    int* number_p = (int*)malloc(sizeof(int) * 1000);
    int* number_q = (int*)malloc(sizeof(int) * 1000);
    preorder_traver(p,number_p,p_length);
    preorder_traver(q,number_q,q_length);
    int i = 0;
    if((*p_length) != (*q_length))
        return false;
    for(i = 0;i < (*p_length);i++)
    {
        if(number_p[i] != number_q[i])
        {
            return false;
        }
    }
    return true;
}

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