Validate Binary Tree中一直出现空指针异常 - Java

Question

Leetcode原题：给一个二叉树，判断这个二叉树是不是二叉搜索树。

Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
A single node tree is a BST

我的思路是用BFS遍历这棵树，然后比较left child，right child是不是符合条件。
但是运行的时候，符合条件的树可以通过，不符合条件的树会一直给我返回空指针异常。
比如这棵树：
{1,#,2,3}
返回：Runtime Error
Exception in thread "main" java.lang.NullPointerException at Solution.isValidBST(Solution.java:35) at Main.main(Main.java:20)

请问为什么会给我报错呢？我只是return false;而已啊？
谢谢。

下面是我的代码：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: True if the binary tree is BST, or false
     */
    public boolean isValidBST(TreeNode root) {
        // write your code here
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        if(root == null){
            return true;
        }
        queue.offer(root);
        while(queue.peek() != null){
            TreeNode parent = queue.poll();
            if(parent.left == null && parent.right == null){
                return true;
            }
            else if(parent.left != null && parent.right == null){
                if(root.val >= root.left.val){
                    queue.offer(parent.left);
                }
                else{
                    return false;
                }
            }
            else if(parent.left == null && parent.right != null){
                if(root.val <= root.right.val){
                    queue.offer(parent.right);
                }
                else{
                    return false;
                }
            }
            else{
                if(root.val >= root.left.val && root.val <= root.right.val){
                    queue.offer(parent.left);
                    queue.offer(parent.right);
                }
                else{
                    return false;
                }
            }
        }
        return false;
    }
}

Answer 1

能否给出构造 {1,#,2,3}这棵树的代码？