有道题不太会，求助大佬们：

在二叉搜索树的基础上，计算每个节点的左子节树的节点数量和右子节点树的节点数量。 如何建立这样的二叉搜索树

我的错误代码（中序输出正确，左右节点数量不对）：

#include<bits/stdc++.h>
using namespace std;

struct Node{
	int key;
	struct Node *left;
	struct Node *right;
	struct Node *parent;
	int result;
	int resultright;
	int resultleft;
};

void makeNode(struct Node *n, int m){
	if(n == 0){
		return;
	}
	
	if(m < n->key){
		if(n->left == 0){
			struct Node *newNode = new struct Node();
			
			newNode->key = m;
			newNode->left = 0;
			newNode->right = 0;
			newNode->parent = n;
			
			n->left = newNode;
			newNode->resultleft++;
		}
		else{
			makeNode(n->left, m);
		}
	}
		
	if(m > n->key){
		if(n->right == 0){
			struct Node *newNode = new struct Node();
			
			newNode->key = m;
			newNode->left = 0;
			newNode->right = 0;
			newNode->parent = n;
			
			n->right = newNode;
			newNode->resultright++;
		}
		else{
			makeNode(n->right, m);
		}
	} 
}

void print(struct Node *n){
	if(n == 0){
		return;
	}
	
	print(n->left);
	cout << n->key << " ";
	print(n->right);
}

struct Node* search(struct Node *n, int t){
	if(n == 0){
		return 0;
	}
	
	if(t == n->key){
		return n;
	}
	else if(t < n->key){
		return search(n->left, t);
	}
	else{
		return search(n->right, t);
	}	
}

int main(){
	struct Node root;
	
	cin >> root.key;
	root.left = 0;
	root.right = 0;
	root.parent = 0;
	
	for(int i = 0;i<7;i++){
		int m;
		cin >> m;
		makeNode(&root, m);
	}
	
	print(&root);
	
	struct Node* p = search(&root, 55);
	if(p != 0){
		cout << p->key << " ";
	}
	
	cout << root.resultright;
	cout << root.resultleft;
	
	return 0;
} 

麻烦大佬们帮我看看

//在线等，急