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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:
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 }
LeetCode Verify Preorder Sequence in Binary Search Tree
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5327638.html
