编写函数，以一颗树的根为输入，返回一颗对应的二叉树，给出详细c++设计代码

为了将树转换为对应的二叉树，我们可以采用先序遍历的方式，将每个节点作为根节点，其所有子节点作为左子树节点，并将右子树节点设为空。具体实现步骤如下：

1. 定义二叉树节点结构体：

typedef struct binary_tree_node {
    int data;
    struct binary_tree_node *left;
    struct binary_tree_node *right;
} BinaryTreeNode;

2. 定义将树转换为二叉树的函数：

BinaryTreeNode* treeToBinaryTree(TreeNode* root) {
    if (root == NULL) {
        return NULL;
    }
    
    BinaryTreeNode *newRoot = (BinaryTreeNode*) malloc(sizeof(BinaryTreeNode));
    newRoot->data = root->val;
    newRoot->left = NULL;
    newRoot->right = NULL;
    
    if (root->children != NULL) {
        newRoot->left = treeToBinaryTree(root->children);
    }
    
    BinaryTreeNode *cur = newRoot->left;
    while (cur != NULL && cur->right != NULL) {
        cur = cur->right;
    }
    
    for (int i = 1; i < root->childrenSize; i++) {
        cur->right = treeToBinaryTree(root->children+i);
        cur = cur->right;
    }
    
    return newRoot;
}

3. 在主函数中调用该函数进行测试：

int main() {
    // 创建一颗树
    TreeNode *root = (TreeNode*) malloc(sizeof(TreeNode));
    root->val = 1;
    TreeNode *node1 = (TreeNode*) malloc(sizeof(TreeNode));
    node1->val = 2;
    TreeNode *node2 = (TreeNode*) malloc(sizeof(TreeNode));
    node2->val = 3;
    TreeNode *node3 = (TreeNode*) malloc(sizeof(TreeNode));
    node3->val = 4;
    TreeNode *node4 = (TreeNode*) malloc(sizeof(TreeNode));
    node4->val = 5;
    
    node1->children = (TreeNode**) malloc(2*sizeof(TreeNode*));
    node1->children[0] = node3;
    node1->children[1] = node4;
    node1->childrenSize = 2;
    
    root->children = (TreeNode**) malloc(2*sizeof(TreeNode*));
    root->children[0] = node1;
    root->children[1] = node2;
    root->childrenSize = 2;
    
    // 将树转换为二叉树
    BinaryTreeNode *newRoot = treeToBinaryTree(root);
    
    return 0;
}