Find the layer number of a node with a value of x in a binary tree

[Problem description]
Taking the binary linked list as the storage structure, write an algorithm to find the layer number of the node with value x in the binary tree.

[Input form] Two lines, the first line is the preorder traversal sequence of the extended binary tree, and the second line is the node x to be queried
[Output form] the node number where the value is x. The layer where the root node is located is recorded as the first layer.
[Sample input] AB # D ## C ##
D
[Sample output]
3
Analysis:
1. Construct a binary tree node
2. Enter the value of the node
3. Traverse the number of record layers
4. Determine the output result

Construct a binary tree node

template <typename DataType>
struct BiNode
{
    DataType data;
    BiNode<DataType>*lchild,*rchild;
};
template<typename DataType>
class BiTree
{
public:
    BiTree()
    {
        root=Create();
    }
    int LevelNum(DataType x)
    {
        return LevelNum(root,1,x);//这里level初始化为1
    }
private:
    BiNode<DataType>*Create();
    int LevelNum(BiNode<DataType>*bt,int level,DataType x);
    BiNode<DataType>*root;

};

Enter node value

template<typename DataType>
BiNode<DataType>*BiTree<DataType>::Create()
{
    BiNode<DataType>*bt;
    char ch;
    cin>>ch;
    if(ch=='#')
        bt=NULL;
    else
    {
        bt=new BiNode<DataType>;
        bt->data=ch;
        bt->lchild=Create();
        bt->rchild=Create();
    }
    return bt;
}

Traverse the record layer

int BiTree<DataType>::LevelNum(BiNode<DataType>*bt,int level,DataType x)
{
    if(bt==NULL)//两种情况 树为空 或者 未找到 x
        return 0;
    else
    {
        if(bt->data==x)//找到x 返回层数
            return level;
        else
        {
            int l1=0;
            l1=LevelNum(bt->lchild,level+1,x);//利用递归 继续寻找左子树 注意层数要增加
            if(l1!=0)
                return l1;
            else
            {
                int l2=0;
                l2=LevelNum(bt->rchild,level+1,x);//利用递归 继续寻找右子树 注意层数要增加
                if(l2!=0)
                    return l2;
                else
                    return 0;//最后没找到 返回0
            }
        }
    }
}

Main function

int main()
{
    BiTree<char>t1;
    char x;
    cin>>x;
    cout<<t1.LevelNum(x);
}

Below is a little white. If there is an inappropriate place, please correct me.