二叉树寻找节点在第几层获取左右子树有几个节点

#include <iostream>
#include <vector>
#include <stdio.h>
using namespace std;

class TreeNode {
public:
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

class Solution {
    vector<int> res;
public:
    void traverse(TreeNode* root,int level) {
        if(root == nullptr) {
            return ;
        }
        cout << "节点" << root->val << "在第" << level << "层" << endl;
        traverse(root->left,level + 1);
        traverse(root->right,level + 1);
    }
    int count(TreeNode* root) {
        if(root == nullptr) {
            return 0;
        }
        int leftCount = count(root->left);
        int rightCount = count(root->right);
        // printf("节点 %s 的左子树有 %d 个节点，右子树有 %d 个节点",root->val, leftCount, rightCount);
        cout << "节点" << root->val << "的左子树有" << leftCount << "个节点，右子树有" << rightCount << "个节点" << endl;
       return leftCount + rightCount + 1;
    }
    
};

TreeNode* newNode(int data)
{
    TreeNode* node = new TreeNode();
    node->val = data;
    node->left = NULL;
    node->right = NULL;

    return(node);
}

int main() {
    /*creating tree*/
    TreeNode *root = newNode(50);
    root->left = newNode(30);
    root->left->left = newNode(20);
    root->left->right = newNode(40);
    root->right = newNode(70);
    root->right->left = newNode(60);
    root->right->right = newNode(80);
//          50
//       /    \
//      30      70
//     /  \    /  \
//   20   40  60   80
    Solution solution;
    solution.traverse(root,1);
    int NodeCount = solution.count(root);
    cout << "一共有" << NodeCount << "个节点" << endl;

    return 0;
}