方法一次只能做一件事。同样，您做事的方式通常也很奇怪。我将给您一些 几乎是Java的伪代码
。抱歉，但是我有一段时间没有接触Java了。希望对您有所帮助。看看我对这个问题也发表的评论，希望您能解决！

像这样调用您的isBST：

public boolean isBst(BNode node)
{
    return isBinarySearchTree(node , Integer.MIN_VALUE , Integer.MIN_VALUE);
}

内部：

public boolean isBinarySearchTree(BNode node , int min , int max)
{
    if(node.data < min || node.data > max)
        return false;
    //Check this node!
    //This algorithm doesn't recurse with null Arguments.
    //When a null is found the method returns true;
    //Look and you will find out.
    /*
     * Checking for Left SubTree
     */
    boolean leftIsBst = false;
    //If the Left Node Exists
    if(node.left != null)
    {
        //and the Left Data are Smaller than the Node Data
        if(node.left.data < node.data)
        {
            //Check if the subtree is Valid as well
            leftIsBst = isBinarySearchTree(node.left , min , node.data);
        }else
        {
            //Else if the Left data are Bigger return false;
            leftIsBst = false;
        }
    }else //if the Left Node Doesn't Exist return true;
    {
        leftIsBst = true;
    }

    /*
     * Checking for Right SubTree - Similar Logic
     */
    boolean rightIsBst = false;
    //If the Right Node Exists
    if(node.right != null)
    {
        //and the Right Data are Bigger (or Equal) than the Node Data
        if(node.right.data >= node.data)
        {
            //Check if the subtree is Valid as well
            rightIsBst = isBinarySearchTree(node.right , node.data+1 , max);
        }else
        {
            //Else if the Right data are Smaller return false;
            rightIsBst = false;
        }
    }else //if the Right Node Doesn't Exist return true;
    {
        rightIsBst = true;
    }

    //if both are true then this means that subtrees are BST too
    return (leftIsBst && rightIsBst);
}

现在：如果要查找每个子树的Min和Max值，则应使用容器（我使用ArrayList），并存储一个Node, Min, Max表示根节点和值的三元组（显然）。

例如。

/*
 * A Class which is used when getting subTrees Values
 */
class TreeValues
{
    BNode root; //Which node those values apply for
    int Min;
    int Max;
    TreeValues(BNode _node , _min , _max)
    {
        root = _node;
        Min = _min;
        Max = _max;
    }
}

还有一个：

/*
 * Use this as your container to store Min and Max of the whole
 */
ArrayList<TreeValues> myValues = new ArrayList<TreeValues>;

现在，这是一种查找给定节点的Min和Max值的方法：

/*
 * Method Used to get Values for one Subtree
 * Returns a TreeValues Object containing that (sub-)trees values
 */ 
public TreeValues GetSubTreeValues(BNode node)
{
    //Keep information on the data of the Subtree's Startnode
    //We gonna need it later
    BNode SubtreeRoot = node;

    //The Min value of a BST Tree exists in the leftmost child
    //and the Max in the RightMost child

    int MinValue = 0;

    //If there is not a Left Child
    if(node.left == null)
    {
        //The Min Value is this node's data
        MinValue = node.data;
    }else
    {
        //Get me the Leftmost Child
        while(node.left != null)
        {
            node = node.left;
        }
        MinValue = node.data;
    }
    //Reset the node to original value
    node = SubtreeRoot; //Edit - fix
    //Similarly for the Right Child.
    if(node.right == null)
    {
        MaxValue = node.data;
    }else
    {
        int MaxValue = 0;
        //Similarly
        while(node.right != null)
        {
            node = node.right;
        }
        MaxValue = node.data;
    }
    //Return the info.
    return new TreeValues(SubtreeRoot , MinValue , MaxValue);   
}

但这仅返回一个节点的值，因此我们将使用它来查找整棵树：

public void GetTreeValues(BNode node)
{
    //Add this node to the Container with Tree Data 
    myValues.add(GetSubTreeValues(node));

    //Get Left Child Values, if it exists ...
    if(node.left != null)
        GetTreeValues(node.left);
    //Similarly.
    if(node.right != null)
        GetTreeValues(node.right);
    //Nothing is returned, we put everything to the myValues container
    return; 
}

使用这种方法，您的通话应该看起来像

if(isBinarySearchTree(root))
    GetTreeValues(root);
//else ... Do Something

这几乎是Java。它应该可以进行一些修改和修复。找到一本很好的面向对象的书，它将对您有所帮助。注意，该解决方案可以分解为更多方法。