问题描述
这是一个程序,用于将数组(其中元素按升序排序)转换为高度平衡的BST.
This is a program for converting an array, where elements are sorted in ascending order, to a height balanced BST.
我输入了五个元素,将它们传递给数组,对数组进行排序并使用方法.
I input five element, pass them to an array, sort the array and use methods.
它会产生此错误:
Exception in thread "main" java.lang.StackOverflowError
at Solution.sortedArrayToBST(Node.java:26)
如何解决此错误?
import java.util.*;
class Node {
int val;
Node left;
Node right;
Node(int x) {
val = x;
}
}
class Solution {
public Node sortedArrayToBST(int[] num) {
if (num.length == 0)
return null;
return sortedArrayToBST(num, 0, num.length - 1);
}
public Node sortedArrayToBST(int[] num, int start, int end) {
int mid = (start + end) / 2;
Node root = new Node(num[mid]);
root.left = sortedArrayToBST(num, start, mid - 1);
root.right = sortedArrayToBST(num, mid + 1, end);
return root;
}
public static void main(String[] args) {
Solution sol = new Solution();
Scanner input = new Scanner(System.in);
int[] numbers = new int[5];
System.out.println("Please enter numbers");
for (int i = 0; i < numbers.length; i++) {
numbers[i] = input.nextInt();
}
// sorting
for (int j = 0; j<numbers.length; j++) {
for (int k = 0; k < numbers.length; k++){
if (numbers[j] < numbers[k]) {
int buffer = numbers[j];
numbers[j] = numbers[k];
numbers[k] = buffer;
}
}
}
sol.sortedArrayToBST(numbers, 0, 5);
sol.sortedArrayToBST(numbers);
}
}
推荐答案
sortedArrayToBST(num,start,end)方法没有终止条件,这就是为什么它永远不会结束的原因,因为您遇到了StackOverFlow错误.您需要在方法的开头添加一个条件:if(start> end)返回null.您可以在将数组排序为BST"问题上检查此Youtube视频:根据排序后的数组创建平衡的二进制搜索树
The method sortedArrayToBST(num, start, end) does not have a termination condition that is why it never ends are you are getting StackOverFlow error.You need to add a condition in the start of the method:if(start > end) return null.You can check this Youtube video on Sorted Array to BST problem:Create a balanced binary search tree from a sorted array
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