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.

一刷
题解：
tree的题目天生就适合用recursion来求解。注意，子问题的函数自带三个参数， root, min, max. 表示子树root内所有的值都在min-max之内。注意，在顶点时，只用Integer.MIN_VALUE和Integer.MAX_VALUE是不够的，因为这样这两个值就被排除在树之外了。所以应该用long型

/**
 * 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) {
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    private boolean isValidBST(TreeNode root, long min, long max){
        if(root == null) return true;
        if(root.val<=min || root.val>=max) return false;
        return isValidBST(root.left, min, root.val) &&
            isValidBST(root.right, root.val, max);
    }
}

二刷
思路与一刷相同

/**
 * 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) {
       return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE); 
    }
    
    public boolean isValidBST(TreeNode root, long min, long max){
        if(root == null) return true;
        if(root.val<=min || root.val>=max) return false;
        return isValidBST(root.left, min, root.val) &&
            isValidBST(root.right, root.val, max);
    }
}