098 Validate Binary Search Tree

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：

Input:
2
/ \
1 3
Output: true

解释下题目：

判断一棵树是否二分搜索树，二分搜索树是左子树都小于根节点，右子树全部大于根子树。

1. 二叉树肯定用递归的啦

实际耗时：0ms

public boolean isValidBST(TreeNode root) {
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }

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

        return isValidBST(root.left, minVal, root.val) && isValidBST(root.right, root.val, maxVal);
    }

  思路就是递归。

时间复杂度O(n) 因为每个节点都遍历一遍

空间复杂度O(1)

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