C语言：重建二叉树

C语言：重建二叉树

学习浙大的数据结构课程做的一些练习。通过前序排序找根节点，在中序排序中通过找到的根节点的值确定此根节点的左子树和右子树。递归实现此过程，每次递归都相当于找到一个新的根节点。

struct TreeNode {
     
	int val;
	struct TreeNode* left;
	struct TreeNode* right;
};

struct TreeNode* buildTree(int* preorder, int preorderSize,
    int* inorder, int inorderSize)
{
     
    if (!(preorderSize && inorderSize))
    {
     
        return NULL;
    }
    int root = preorder[0];
    int i = 0;
    for (i = 0; i < inorderSize; i++)
    {
     
        if (inorder[i] == root)
        {
     
            break;
        }
    }
    int newlsize = i;
    int newrsize = preorderSize - i -1;

    int* newleftpre = malloc(newlsize * sizeof(int));


    for (int j = 0; j < newlsize; j++)
    {
     
        newleftpre[j] = preorder[j + 1];
    }

    int* newleftin = malloc(newlsize * sizeof(int));

    for (int j = 0; j < newlsize; j++)
    {
     
        newleftin[j] = inorder[j ];
    }

    int* newrightpre = malloc(newrsize * sizeof(int));

    for (int j = 0; j < newrsize; j++)
    {
     
        newrightpre[j] = preorder[j+newlsize+1];
    }

    int* newrightin = malloc(newrsize * sizeof(int));

    for (int j = 0; j < newrsize; j++)
    {
     
        newrightin[j] = inorder[j + newlsize + 1];
    }

    struct TreeNode* tmp = malloc(sizeof(struct TreeNode));
    tmp->val = root;
    tmp->left = buildTree(newleftpre, newlsize,
        newleftin, newlsize);
    tmp->right = buildTree(newrightpre, newrsize,
        newrightin, newrsize);

    free(newleftpre);
    free(newleftin);
    free(newrightpre);
    free(newrightin);

    return tmp;



}


int main()
{
     

    int pre[7] = {
      3, 9, 5, 13, 20, 15, 7 };
    int in[7] = {
      5,13,9,3,15,20,7 };


    struct TreeNode* tr = NULL;

    tr = buildTree(pre, 7, in, 7);

    printf("%d\t%d\t%d\t%d\n", tr->val, tr->left->val,
        tr->left->left->val,tr->left->left->right->val);

	//最后要记得遍历二叉树释放内存，这里省略


	return EXIT_SUCCESS;
}