Determine whether there is a node with value x in the binary tree

Taking the binary linked list as the storage structure, write an algorithm to judge whether there is a 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] If the node x exists, output "YES"; otherwise output "NO".
[Sample input] AB # D ## C ##
D
[Sample output]
YES
analysis:
1. Construct a binary tree node
2. Enter the node value
3. Traverse
4. Judge the output result

Construct a binary tree node

template <typename DataType>
struct BiNode
{
	DataType data;
	BiNode<DataType>*rchild,*lchild;
}
class BiTree
{
	public:
	BiTree()
	{
	 root=Create;
	}
	bool ExistX(DataType x)
	return ExistX(root,x);
	private:
	BiNode<DataType>*root;
	BiNode<DataType>*Create();
	bool ExistX(BiNode<DataType>*bt,DataType x);
}

Use function to
input node value

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

Traversal detection function

template <typename DataType>
bool BiTree<DataType>::ExistX(BiNode<DataType>*bt,DataType x)
{
    if(bt==NULL)//如果为空 未查到 false
        return false;
    else
    {
        if(bt->data==x)//查找的 true
        {
            return true;
        }
        else
        {
            if(ExistX(bt->lchild,x))
                return true;//如果在左孩子找到return true
            else			
            {
                if(ExistX(bt->rchild,x))//在右孩子找到 reture true
                    return true;
                else
                    return false;//都没找到 return false
            }

        }
    }



}

Main function

int main()
{
    BiTree<char>t1;
    char x;
    cin>>x;
    if(t1.ExistX(x))
        cout<<"YES";
    else
        cout<<"NO";
}

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