Validate Binary Search Tree

https://leetcode.com/problems/validate-binary-search-tree/

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.

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}”.

题意：检查一颗二叉树是否为二叉搜索树。
ps:二叉查找树（Binary Search Tree），（又：二叉搜索树，二叉排序树）它或者是一棵空树，或者是具备下列性质的二叉树： 若它的左子树不空，则左子树上全部结点的值均小于它的根结点的值； 若它的右子树不空，则右子树上全部结点的值均大于它的根结点的值； 它的左、右子树也分别为二叉排序树。
思路：DFS遍历二叉树，检索左子树时以父节点下限为下限，已父节点为上限；检索右子树时以父节点为下限，以父节点上限为上限。根节点上限下限分别为int的上下限。
实现：

public class Solution {

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

      public boolean isValidBST(TreeNode root , long min, long max ) {
           if (root == null)
               return true ;
           if (root .val > min && root. val < max)
               return isValidBST(root .left , min , root .val )
                        && isValidBST( root. right, root .val , max );
           return false ;

     }
}