validate-binary-search-tree Java code

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.

confused what”{1,#,2,3}”means? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

1

/ \

2 3

/

4

\

5

The above binary tree is serialized as”{1,2,3,#,#,4,#,#,5}”.

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public static boolean IsSubtreeLessThan(TreeNode t, int val) {
if (t == null)
return true;
return (t.val < val && IsSubtreeLessThan(t.left, val) && IsSubtreeLessThan(t.right, val));
}
public static boolean IsSubtreeMoreThan(TreeNode t, int val) {
if (t==null)
return true;
return (t.val>val && IsSubtreeMoreThan(t.left, val) && IsSubtreeMoreThan(t.right, val));
}

public boolean isValidBST(TreeNode root) {
if (root == null)
return true;
return (IsSubtreeLessThan(root.left, root.val) && IsSubtreeMoreThan(root.right, root.val)
&& isValidBST(root.left) && isValidBST(root.right));
}
}