Leetcode - 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.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

[分析] 这题较一年之前的版本多了些边界值的test case，看出leetcode在不断优化完善，赞!
对我来说，对边界条件的考虑是bug free的关键。基本思路，判断根节点是否满足条件，然后递归判断左右子节点是否满足条件。判断是否满足BST性质，可以从根节点下传节点的值域，root的初始值域是多少呢，(Integer.MIN_VALUE, Integer.MAX_VALUE)貌似是一个合理的选择，但如果root本身就是Integer.MIN_VALUE或者是Intger.MAX_VALUE呢？ 因此还需要使用两个辅助flag，effLowBound, effUpBound分别表示值域的上下边界是否有效，对于root，上下边界均无效。

public class Solution {
    public boolean isValidBST(TreeNode root) {
        return check(root, Integer.MIN_VALUE, Integer.MAX_VALUE, false, false);
    }
    public boolean check(TreeNode curr, int low, int up, 
            boolean effLowBound, boolean effUpBound) {
        if (curr == null) return true;
        if ((effLowBound && curr.val <= low) || (effUpBound && curr.val >= up))
            return false;
        return check(curr.left, low, curr.val, effLowBound, true)
                && check(curr.right, curr.val, up, true, effUpBound);
    }
}