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?

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/* Stack time O(n)space O(n)
public class Solution {
    public boolean verifyPreorder(int[] preorder) {
        //stack preorder root -left - right  先降序后升序
        Stack<Integer> stack = new Stack<>();
        int min = Integer.MIN_VALUE;
        for(int i = 0; i < preorder.length; i ++){
            if(preorder[i] < min)  
                return false;
            while(!stack.isEmpty() && stack.peek() < preorder[i]){
                min = stack.pop();
            }
            stack.push(preorder[i]);
        }
        return true;
    }
}*/

public class Solution {
    public boolean verifyPreorder(int[] preorder) {
        int i = -1;
        int min = Integer.MIN_VALUE;
        for(int num : preorder){
            if(num < min)
                return false;
            while(i >= 0 &&  preorder[i] < num){
                min = preorder[i--];
            }
            preorder[++i] = num;
        }
        return true;
    }
}

Follow up 参考 https://segmentfault.com/a/1190000003874375

如何验证中序序列？A：中序序列是有序的，只要验证其是否升序的就行了。

如何验证后序序列？

后序序列的顺序是left - right - root，而先序的顺序是root - left - right。我们同样可以用本题的方法解，不过是从数组的后面向前面遍历，因为root在后面了。而且因为从后往前看是先遇到right再遇到left，所以我们要记录的是限定的最大值，而不再是最小值，栈pop的条件也变成pop所有比当前数大得数。栈的增长方向也是从高向低了。

public static boolean verifyPostorder(int[] preorder) {
            int i = preorder.length;
            int max = Integer.MAX_VALUE;
            for(int j = preorder.length -1; j >=0; j--){
                if(preorder[j] > max)
                    return false;
                while(i < preorder.length &&  preorder[i] < preorder[j]){
                    max = preorder[i++];
                }
                preorder[--i] = preorder[j];
            }
            return true;
        }