255. Verify Preorder Sequence in Binary Search Tree

Given an array of numbers, verify whether it is the correct preorder traversal sequence of a binary search tree.

You may assume each number in the sequence is unique.

Follow up:
Could you do it using only constant space complexity?

要验证二叉搜索树是不是合法，核心是左节点比根节点小，右节点比根节点大。这里stack只保存左树，如果当前值大于peek，说明到了一个分叉，该访问右子树，low更新为peek，下面的数都不能比这个low小，否则返回false。代码如下：

public class Solution {    public boolean verifyPreorder(int[] preorder) {        int low = Integer.MIN_VALUE;        Stack<Integer> stack = new Stack<Integer>();        for (int val: preorder) {            if (val < low) {                return false;            }            while (!stack.isEmpty() && val > stack.peek()) {                low = stack.pop();            }            stack.push(val);        }        return true;    }}