LeetCode 255. Verify Preorder Sequence in Binary Search Tree

原题链接在这里：https://leetcode.com/problems/verify-preorder-sequence-in-binary-search-tree/

题目：

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?

题解：

Method 1 利用stack模拟preorder. stack中放的是左子树. 遇到比stack top还要大, 说明遇到了以top 为root的右子树，把左子树连 & root 都pop出来. 

low是当前的lower bound, pop出来说明左边扫完了，不可能出现更小的了，所以更新low. 若是遇到比low还小就说明不是preorder.

Time Complexity: O(n). Space: O(logn).

 

Method 2 是在array本身上实现stack的功能.

index is initialized as -1, and assign the top of stack as preoder[++index]. But not initialize index as 0, and assignthe top of stack as preorder[index++].

Since every time, the top of stack is preorder[index]. If use index++, it is not the top of stack. Also there is chance of OutOfIndexException.

Time Complexity: O(n). Space: O(1).

AC Java:

 1 public class Solution {
 2     public boolean verifyPreorder(int[] preorder) {
 3         /*
 4         //Method 1
 5         int low = Integer.MIN_VALUE;
 6         Stack<Integer> stk = new Stack<Integer>();
 7         for(int num : preorder){
 8             if(num < low){
 9                 return false;
10             }
11             while(!stk.isEmpty() && stk.peek() < num){
12                low = stk.pop();
13             }
14             stk.push(num);
15         }
16         return true;
17         */
18         int index = -1;
19         int low = Integer.MIN_VALUE;
20         for(int num : preorder){
21             if(num < low){
22                 return false;
23             }
24             while(index >= 0 && preorder[index] < num){
25                 low = preorder[index--];
26             }
27             preorder[++index] = num;
28         }
29         return true;
30     }
31 }