题解 | #判断一棵二叉树是否为搜索二叉树和完全二叉树#

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

/**
 * NC60 判断一棵二叉树是否为搜索二叉树和完全二叉树
 * @author d3y1
 */
public class Solution {
    private TreeNode pre = null;
    private boolean[] result = new boolean[]{true, true};
    private boolean reachEnd = false;

    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 程序入口
     *
     * @param root TreeNode类 the root
     * @return bool布尔型一维数组
     */
    public boolean[] judgeIt (TreeNode root) {
        inorder(root);
        levelorder(root);
        return result;
    }

    /**
     * 递归: 中序遍历
     * @param root
     */
    private void inorder(TreeNode root){
        if(root == null){
            return;
        }

        inorder(root.left);

        if(pre != null){
            if(pre.val > root.val){
                result[0] = false;
            }
        }
        pre = new TreeNode(root.val);

        inorder(root.right);
    }

    /**
     * 层序遍历: 队列
     * 完全二叉树在遇到空节点之后剩余的应当全是空节点
     * @param root
     */
    private void levelorder(TreeNode root){
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        TreeNode node;
        while(queue.peek() != null){
            node = queue.poll();
            queue.offer(node.left);
            queue.offer(node.right);
        }

        // 剩余的应当全是空节点
        while(!queue.isEmpty() && queue.peek()==null){
            queue.poll();
        }

        if(!queue.isEmpty()){
            result[1] = false;
        }
    }

    /**
     * 层序遍历: 队列
     * @param root
     */
    private void levelorder1(TreeNode root){
        if(root == null){
            return;
        }

        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        TreeNode node;
        while(!queue.isEmpty()){
            node = queue.poll();
            if(reachEnd){
                if(node.left!=null || node.right!=null){
                    result[1] = false;
                    break;
                }
            }else{
                if(node.left==null && node.right==null){
                    reachEnd = true;
                }else if(node.left==null && node.right!=null){
                    result[1] = false;
                    break;
                }else if(node.left!=null && node.right==null){
                    reachEnd = true;
                    queue.offer(node.left);
                }else{
                    queue.offer(node.left);
                    queue.offer(node.right);
                }
            }
        }
    }
}