将排序后的数组转换为二进制搜索树(Java)堆栈溢出异常

本文介绍了将排序后的数组转换为二进制搜索树(Java)堆栈溢出异常的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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