255. Verify Preorder Sequence in Binary Search Tre

/*
 * 255. Verify Preorder Sequence in Binary Search Tree 
Total Accepted: 8634 Total Submissions: 23097 Difficulty: Medium
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?

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 */
package tree;

import java.util.Stack;

public class VerifyPreOrderBST {
    public boolean verifyPreorder(int[] preorder) {
        // store the nodes using a stack path.  if path.isEmpty() or p < path.peek(), we still in the left subtree, just push p into path;  if p > path.peek(), we pop the node as a low bound, push p into path.  For the next p if smaller than low bound. return false, since being in their right subtree means we must never come across a smaller number anymore.
        // https://leetcode.com/discuss/51543/java-o-n-and-o-1-extra-space
        // test case : [4, 2, 1, 3, 6, 8,5] => false
        if (preorder == null || preorder.length == 0) {
            return true;
        }
        int low = Integer.MIN_VALUE;
        Stack<Integer> path = new Stack<>();
       for (int p : preorder) {
           if (p < low) {
               return false;
           }
           while (!path.isEmpty() && p > path.peek()) {
               low = path.pop();
           }
           path.push(p);
       }
       return true;
    }
}