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.
A single node tree is a BST

分析：此题需要注意的是一个平衡二叉树的条件为左子树的val中最大的值需要小于root.val，右子树的val最小值需要大于root.val，一旦有任意子树不满足上述条件，则该树不是平衡二叉树。

class ResultType {    boolean isValid;    int max;    int min;    ResultType(boolean a, int b, int c) {        isValid = a;        max = b;        min = c;    }}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) {        ResultType rt = helper(root);        return rt.isValid;    }    private ResultType helper(TreeNode root) {        if (root == null) {            return new ResultType(true, Integer.MIN_VALUE, Integer.MAX_VALUE);        }        ResultType left = helper(root.left);        ResultType right = helper(root.right);        if (!left.isValid || !right.isValid) {            return new ResultType(false, 0, 0);        }        if ((root.left != null && left.max >= root.val) || (root.right != null && right.min <= root.val)) {            return new ResultType(false, 0, 0);        }        return new ResultType(true,                              Math.max(root.val, right.max),                              Math.min(root.val, left.min));    }}