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?
链接： http://leetcode.com/problems/verify-preorder-sequence-in-binary-search-tree/
题解：
使用Stack来模拟preorder traversal -> mid, left, right。主要代码都是参考了Stefan Pochmann的解答。
Time Complexity - O(n)， Space Complexity - O(logn)
public class Solution {
public boolean verifyPreorder(int[] preorder) {
int low = Integer.MIN_VALUE;
Stack<Integer> stack = new Stack<>();
for(int i : preorder) {
if(i < low)
return false;
while(!stack.isEmpty() && i > stack.peek())
low = stack.pop();
stack.push(i);
}
return true;
}
}
不使用Stack，Space Complexity O(1)的解法， 利用了原数组
public class Solution {
public boolean verifyPreorder(int[] preorder) {
int low = Integer.MIN_VALUE, index = -1;
for(int i : preorder) {
if(i < low)
return false;
while(index >= 0 && i > preorder[index])
low = preorder[index--];
preorder[++index] = i;
}
return true;
}
}
Reference:
https://leetcode.com/discuss/51543/java-o-n-and-o-1-extra-space
https://leetcode.com/discuss/52060/72ms-c-solution-using-one-stack-o-n-time-and-space
https://leetcode.com/discuss/65241/ac-python-o-n-time-o-1-extra-space
https://leetcode.com/discuss/68862/my-c-solution-easy-to-understand