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.

Solution1: recursion
中序遍历结果是按顺序递增的。根据这一点我们只需要中序遍历这棵树，然后保存前驱结点，每次检测是否满足递增关系即可。注意以下代码没用一个变量去保存前驱结点，原因是java没有传引用的概念，如果传入一个变量，它是按值传递的，所以是一个备份的变量，改变它的值并不能影响它在函数外部的值，算是java中的一个小细节。
做一次树的遍历，所以时间复杂度是O(n)，空间复杂度是O(logn)。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        ArrayList res = new ArrayList(); //store value of last node
        res.add(null);
        return helper(root, res);
    }
    
    public boolean helper(TreeNode root, ArrayList res) {
        if(root == null)
            return true;
        
        boolean left = helper(root.left, res);
        
        //in BST, the result of inorder traversal is in ascending order 
        if(res.get(0) != null && res.get(0) >= root.val) 
            return false;

        res.set(0, root.val);
        boolean right = helper(root.right, res);
        
        return left && right;
    }
}

Solution2: recursion

O(n) runtime, O(logn) stack space – Top-down recursion

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        return helper(root, null, null);
    }
    
    public boolean helper(TreeNode root, Integer min, Integer max) {
        if(root == null)
            return true;
        
        return (min == null || root.val > min) && (max == null || root.val < max)
               && helper(root.left, min, root.val) && helper(root.right, root.val, max);
    }
}