Zenefits Interview - Can represent the pre-order list of a BST

Given a list of numbers, determine whether it can represent the pre-order traversal list of a binary search tree (BST).

 

Input

The first line contains the number of test cases, T. T lines follow, consisting of two lines each.

The first line of each test case contains the number of nodes in the tree, N. In next line there will a list of N unique numbers, where each number is from the range [1, N].

 

Output

For each test case, print the string “YES” if there exists a BST whose pre-order traversal is equal to the list, otherwise print the string “NO” (without quotes, preserving capitalization).

 

Constraints

1 ≤ T ≤ 10

1 ≤ N ≤ 100

 

Sample Input

5

3

1 2 3

3

2 1 3

6

3 2 1 5 4 6

4

1 3 4 2

5

3 4 5 1 2

 

Sample Output

YES

YES

YES

NO

NO

 

Explanation

• The first three cases are from the above examples.
• In case 4, after encountering the 3, the 4tells us we are on the right sub-tree, which means no values smaller than 3 are allowed any longer. So when we see the 2 we know the list is invalid.
• Similarly, in case 5, after encountering the 3, the 4 and 5 tell us we are on the right sub-tree, so the subsequent encounter of values 2 and 1, which belong in the left sub-tree, tells us that the list is not valid as a pre-order traversal of a BST.

public static boolean isValidPreorderBST(int[] A) {
	return isValidPreorderBST(A, 0, A.length-1);
}
public static boolean isValidPreorderBST(int[] A, int start, int end) {
	if(start >= end) return true;
	int i=start+1;
	for(; i<=end; i++) {
		if(A[i] > A[start]) break;
	}
	int j = i;
	while(j <= end) {
		if(A[j++] <= A[start]) return false;
	}
	return isValidPreorderBST(A, start+1, i-1) && isValidPreorderBST(A, i, end);
}

public static void main(String[] args) {
	String[] cases = new String[]{"1 2 3", "2 1 3", "3 2 1 5 4 6", "1 3 4 2", "3 4 5 1 2"};
	for(String s:cases) {
		String[] nums = s.split("\\s");
		int[] A = new int[nums.length];
		for(int i=0; i<A.length; i++) {
			A[i] = Integer.parseInt(nums[i]);
		}
		System.out.println(isValidPreorderBST(A));
	}
	
}

 

重构了一下，添加了min数组，保存从当前位置往后的最小值，省去了继续搜索比根节点值小的元素的时间。

public static boolean isValidPreorderBST(int[] A) {
	int n = A.length;
	if(n <= 1) return true;
	int[] min = new int[n];
	min[n-1] = A[n-1];
	for(int i=n-2; i>=0; i--) {
		min[i] = Math.min(min[i+1], A[i]);
	}
	return isValidPreorderBST(A, min, 0, n-1);
}
private static boolean isValidPreorderBST(int[] A, int[] min, int start, int end) {
	if(start >= end) return true;
	int i=start+1;
	for(; i<=end; i++) {
		if(A[i] > A[start]) break;
	}
	if(i<end && min[i+1] <= A[start]) return false;
	return isValidPreorderBST(A, min, start+1, i-1) && isValidPreorderBST(A, min, i, end);
}

 

以上解法的时间复杂度都是O(N^2)。

以下链接提供了O(N)的解法。

http://www.geeksforgeeks.org/construct-bst-from-given-preorder-traversa/