Niuke.com, to determine whether the binary tree is a complete binary tree and BST

Title description

solution:

To judge BST, it is necessary: ​​min<current node<max, and its left and right subtrees also meet similar rules.
To judge a complete binary tree, you need to judge according to the layer, as long as there is one node:

if(cur.left==null && cur.right!=null) return false;

It is not a complete binary tree.
See code:

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    
    
    /**
     * 
     * @param root TreeNode类 the root
     * @return bool布尔型一维数组
     */
    public boolean[] judgeIt (TreeNode root) {
    
    
        
        // write code here
        boolean[]res = new boolean[2];
        if(root==null) {
    
    
            res[0] = true;
            res[1] = true;
            return res;
        }
        res[0] = isBST(root,Integer.MAX_VALUE,Integer.MIN_VALUE);
        res[1] = isBinaryTree(root);
        return res;
    }
    //判断BST二叉树，使用一个max和min来限制范围：
    //当前节点必须满足： max>=root>=min
    public boolean isBST(TreeNode root,int max,int min){
    
    
        if(root== null) return true;
        if(root.val>=max || root.val<=min){
    
    
            return false;
        }
        
        boolean leftRes = isBST(root.left,root.val,min);
        boolean rightRes = isBST(root.right,max,root.val);
        return leftRes && rightRes;
        
    }
    public boolean isBinaryTree(TreeNode root){
    
    
        if(root==null) return true;
        //完全二叉树的标号同满二叉树的标号则相同。
        //按照层级遍历的序号，如果得到的数据和满二叉树的节点值一致则为true
        int fullId = 0;//满二叉树的id
        int judgeId = 0;//完全二叉树的id
        LinkedList<TreeNode> que = new LinkedList<>();
        que.add(root);
        while(!que.isEmpty()){
    
    
            int size = que.size();
            for(int i=0;i<size;i++){
    
    //当前层的节点
                TreeNode cur = que.remove();
                if(cur!=null){
    
    
                    if(cur.left==null && cur.right!=null) return false;
                    
                    que.add(cur.left);
                    que.add(cur.right);
                }else continue;
               
                
            }
       }
        return true; 
    }
}