java 验证二叉搜索树

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:

Input:
    2
   / \
  1   3
Output: true

Example 2:

    5
   / \
  1   4
     / \
    3   6
Output: false
Explanation: The input is: [5,1,4,null,null,3,6]. The root node's value
             is 5 but its right child's value is 4.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
// 解题核心： 验证是否是有效的二叉搜索树 等价于 中序遍历二叉树 且 保证遍历的结果为 升序 即可
// 中序遍历： 
//       inOrder(root.left);
//       println(root.val);
//       inOrder(root.right);
class Solution {
    // 保存遍历的前一个数 方便与当前数进行比较
    // Double.MIN_VALUE 是正数，一个无限接近零的数，Integer.MIN_VALUE才是一个负数
    double prv = -Double.MAX_VALUE;
    public boolean isValidBST(TreeNode root) {
        if(root==null)
            return true;
        if(isValidBST(root.left))   // 先遍历左节点
        {
            if(prv<root.val)        // 这一步就是中序遍历的输出：只不过现在我们要先比较是否
            {                       // 当前是的值比前一个要大，如果小于的话，直接则不需要继
                prv = root.val;     // 续递归
                return isValidBST(root.right);  // 遍历右子树
            }
        }
        return false;
    }
}