255. Verify Preorder Sequence in Binary Search Tree

Description

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?

Solution

DFS, time O(n * log n), space O(1)

这道题在递归到子问题之前，需要做剪枝，否则有些test case会TLE。

class Solution {
    public boolean verifyPreorder(int[] preorder) {
        return verify(preorder, 0, preorder.length - 1);
    }
    
    public boolean verify(int[] preorder, int start, int end) {
        if (start > end) {
            return true;
        }
        
        int pivot = preorder[start];
        int rightStart = -1;     // start index of right subtree

        for (int i = start + 1; i <= end; ++i) {
            if (rightStart == -1 && preorder[i] > pivot) {
                rightStart = i;
            } else if (rightStart != -1 && preorder[i] < pivot) {
                return false;    // prune!
            }
        }
        
        return rightStart == -1 
            ? verify(preorder, start + 1, end)
            : verify(preorder, start + 1, rightStart - 1) 
                && verify(preorder, rightStart, end);
    }
}

Stack, time O(n), space O(n)

非常有趣的写法，值得深思...

class Solution {
    public boolean verifyPreorder(int[] preorder) {
        int min = Integer.MIN_VALUE;
        Stack stack = new Stack<>();
        
        for (int val : preorder) {
            if (val < min) {
                return false;
            }
            while (!stack.empty() && stack.peek() < val) {  // enters right subtree
                min = stack.pop();  // pop nodes in the left of val and update min
            }
            stack.push(val);
        }
        
        return true;
    }
}