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.

Example 1:

  2
 / \
1   3

Binary tree [2,1,3], return true.

Example 2:

  1
 / \
2   3

Binary tree [1,2,3], return false.

这道题看的答案，第一次用到这种限制max, min bound的方法，很巧妙。注意，判断左右子树的时候，记得要更新max, min bound传递到递归里。留意一下，像Integer.MAX_VALUE，Integer.MIN_VALUE一样，double也有类似的Double.NEGATIVE_INFINITY, Double.POSITIVE_INFINITY.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null){
            return true;
        } 
        return helper(root, Double.NEGATIVE_INFINITY, Double.POSITIVE_INFINITY);  
    }
    
    private boolean helper(TreeNode root, double min, double max){
        if (root.val <= min || root.val >= max){
            return false;
        }
        if (root.left != null && !helper(root.left, min, root.val)){
            return false;
        }
        if (root.right != null && !helper(root.right, root.val, max)){
            return false;
        }
        return true;
    }
}

这道题还有另外一个方法，不过要想到的话要求你对BST的性质要比较熟悉。这个方法利用到了BST对应节点的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这个特点，对BST进行inorder遍历，遍历出来的节点val值就是从小到大排列的。如果出现了list.get(i) >= list.get(i + 1)的情况，就说明不是BST.顺便也复习了一下inorder的recursive的做法。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null){
            return true;
        }
        List<Integer> list = new ArrayList<>();
        list = inOrder(root, list);
        for (int i = 0; i < list.size() - 1; i++){
            if (list.get(i) >= list.get(i + 1)){
                return false;
            }
        }  
        return true;
    }
    
    private List<Integer> inOrder(TreeNode root, List<Integer> list){
        if (root.left != null){
            inOrder(root.left, list);
        }
        list.add(root.val);
        if (root.right != null){
            inOrder(root.right, list);
        }
        return list;
    }
}