**[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;    }    }