255. 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?

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

巴特西

255. Verify Preorder Sequence in Binary Search Tree

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