7-3 是否完全二叉搜索树

将一系列给定数字顺序插入一个初始为空的二叉搜索树（定义为左子树键值大，右子树键值小），你需要判断最后的树是否一棵完全二叉树，并且给出其层序遍历的结果

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;

struct Node {
    int val, depth;
    Node *left;
    Node *right;
    Node() { val = 0; depth = 0; left = right = NULL; }
};

struct BST {
    Node *root;
    int cnt, inorder[25], maxdep;
    void init() { root = new Node; cnt = 0; maxdep = 0; }
    void insert(int x) {
        Node *t = root, *fa = root;
        int depth = 0;
        while (t != NULL) {
            depth++;
            fa = t;
            if (x > (t->val)) t = (t->left);
            else t = (t->right);
        }
        t = new Node;
        if (x > fa->val) fa->left = t;
        else fa->right = t;
        t->val = x;
        t->depth = depth;
        maxdep = max(maxdep, depth);
    }
    void bfs() {
        bool ok = true;
        queue<Node*> q;
        q.push(root->left);
        bool flag = false;
        while (!q.empty()) {
            Node *t = q.front(); q.pop();
            inorder[++cnt] = t->val;
            if (t->depth == maxdep) continue;
            if (t->depth < maxdep - 1) {
                if (t->left != NULL) q.push(t->left);
                else ok = false;
                if (t->right != NULL) q.push(t->right);
                else ok = false;
            }
            else if (t->depth == maxdep - 1) {
                if (t->left != NULL && t->right != NULL) {
                    q.push(t->left); q.push(t->right);
                    if (flag) ok = false;
                }
                else if (t->left != NULL && t->right == NULL) {
                    if (flag) ok = false;
                    flag = true; q.push(t->left);
                }
                else if (t->left == NULL && t->right != NULL) { flag = true; ok = false; q.push(t->right); }
                else flag = true;
            }
        }
        printf("%d", inorder[1]);
        for (int i = 2; i <= cnt; i++) printf(" %d", inorder[i]);
        printf("\n");
        if (ok) printf("YES\n");
        else printf("NO\n");
    }
}bst;

int main() {
    int n, t;
    bst.init();
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &t);
        bst.insert(t);
    }
    bst.bfs();
    return 0;
}