**[Lintcode]Validate Binary Search Tree 验证二叉查找树

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

Example

An example:

  2
 / \
1   4
   / \
  3   5

The above binary tree is serialized as {2,1,4,#,#,3,5} (in level order).

分析：首先，每一级可以验证其left和right节点，但是这样做可能会出现left的right大于root的情况，对于这种隔级比较的问题，可以换个思路，每次递归时，传递当前节点的上限/下限。

/**
 * 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) {  
        return helper(root, Integer.MIN_VALUE, Integer.MAX_VALUE);
    }
    
    boolean helper(TreeNode root, int minLimit, int maxLimit) {
        if(root == null) return true;
        
        if((minLimit != Integer.MIN_VALUE ? root.val > minLimit : true) &&
            (maxLimit != Integer.MAX_VALUE ? root.val < maxLimit : true) &&
            helper(root.left, minLimit, root.val)
            && helper(root.right, root.val, maxLimit)) 
            return true;
        else 
            return false;
    }
    
}