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.

tag: binary tree

自己的解法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private class JieTree{
        long minVal;
        long maxVal;
        boolean isBST;

        public JieTree(long minVal, long maxVal, boolean isBST){
            this.minVal = minVal;
            this.maxVal = maxVal;
            this.isBST = isBST;
        }
    }

    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        return helper(root).isBST;
    }

    private JieTree helper(TreeNode root){
        if (root == null){
            return new JieTree(Long.MAX_VALUE, Long.MIN_VALUE, true);
        }

        if (root.left == null && root.right == null){
            return new JieTree(root.val, root.val, true);
        }

        JieTree left = helper(root.left);
        JieTree right = helper(root.right);

        long curMin = Math.min(root.val, left.minVal);
        long curMax = Math.max(root.val, right.maxVal);
        boolean isBST = left.maxVal < root.val && right.minVal > root.val && left.isBST && right.isBST;

        return new JieTree(curMin, curMax, isBST);
    }
}