LeetCode 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，把左子树连同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.

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

AC Java:

public class Solution {
public boolean verifyPreorder(int[] preorder) {
/*
//Method 1
int low = Integer.MIN_VALUE;
Stack<Integer> stk = new Stack<Integer>();
for(int num : preorder){
if(num < low){
return false;
}
while(!stk.isEmpty() && stk.peek() < num){
low = stk.pop();
}
stk.push(num);
}
return true;
*/
int index = -1;
int low = Integer.MIN_VALUE;
for(int num : preorder){
if(num < low){
return false;
}
while(index >= 0 && preorder[index] < num){
low = preorder[index--];
}
preorder[++index] = num;
}
return true;
}
}