题解 | #判断是不是二叉搜索树 分治思想#

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param root TreeNode类 
     * @return bool布尔型
     */
    public boolean isValidBST (TreeNode root) {
        // write code here
        if(root == null){
            return true;
        }
        return getMaxLeftNode(root.left) < root.val &&
                getMinRightNode(root.right) > root.val &&
                isValidBST(root.left) &&
                isValidBST(root.right);
    }

    public static int getMaxLeftNode(TreeNode node){
        if (node == null){
            return Integer.MIN_VALUE;
        }
        while (node.right != null){
            node = node.right;
        }
        return node.val;
    }
    public static int getMinRightNode(TreeNode node){
        if (node == null){
            return Integer.MAX_VALUE;
        }
        while (node.left != null){
            node = node.left;
        }
        return node.val;
    }
}