98. 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.

Example 1:

    2
   / \
  1   3

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

Example 2:

    1
   / \
  2   3

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

##Approach 1: iterative 解题思路为inorder traversal。

由于是BST，其inorder traversal是排好序的，因此可以采用inorder traversal，每次都与前一个节点比较。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        Stack<TreeNode> stack = new Stack<>();
        TreeNode cur = root;
        TreeNode pre = null;
        while(!stack.isEmpty() || cur != null) {
            if(cur != null) {
                stack.push(cur);
                cur = cur.left;
            } else {
                TreeNode node = stack.pop();
                if(pre != null && pre.val >= node.val) return false;
                pre = node;
                cur = node.right;
            }
        }
        return true;
    }
}

##Approach 2: recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isValidBST(TreeNode root) {
        if(root == null) return true;
        return isBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    private boolean isBST(TreeNode root, long min, long max) {
        if(root == null) return true;
        if(root.val <= min || root.val >= max) return false;
        return isBST(root.left, min, root.val) && isBST(root.right, root.val, max);
    }
}