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.

难度:medium

题目:给定二叉树,判断其是否为二叉搜索树。

思路:递归,并记录当前结点的取值范围。

Runtime: 0 ms, faster than 100.00% of Java online submissions for Validate Binary Search Tree.
Memory Usage: 37.6 MB, less than 100.00% of Java online submissions for Validate Binary Search Tree.

/**
 * 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 (null == root) {
            return true;
        }
        
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }

    public boolean isValidBST(TreeNode root, long minVal, long maxVal) {
        if (null == root) {
            return true;
        }
        
        if (root.val >= maxVal || root.val <= minVal) {
            return false;
        }
        
        return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
    }
}

linm
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