三个主要方法的模板
import java.util.*; public class Main { // main public static void main(String[] args) { Scanner sc = new Scanner(System.in); int tree[] = new int[1000]; int arr[] = { 0, 1, 3, 5, 7, 9, 11 }; // for (int i = 1; i < length; i++) { //存储数据 // arr[i] = sc.nextInt(); // } create_tree(arr, tree, 0, 1, arr.length - 1); // 这里记得减1 因为arr[]是从0开始存储数据 for (int i = 0; i < tree.length; i++) { System.out.print(tree[i] + " "); } System.out.println(); update(arr, tree, 0, 1, arr.length - 1, 5, 6); for (int i = 0; i < tree.length; i++) { System.out.print(tree[i] + " "); }System.out.println(); System.out.println(querey(tree, arr, 0, 1, 6, 2, 5)); } // create public static void create_tree(int arr[], int tree[], int node, int start, int end) { // create建树方法!!! if (start == end) { // 离散化的点进行调整!!! tree[node] = arr[start]; } else { int left_node = 2 * node + 1; // 0 | 1 2 | 3 4 5 6 ........... int right_node = 2 * node + 2; int mid = (start + end) / 2; create_tree(arr, tree, left_node, start, mid); create_tree(arr, tree, right_node, mid + 1, end); tree[node] = tree[left_node] + tree[right_node]; // 核心代码 } } // update public static void update(int arr[], int tree[], int node, int start, int end, int idx, int val) { if (start == end) { arr[start] = val; tree[node] = val; } else { int left_node = 2 * node + 1; // 0 | 1 2 | 3 4 5 6 ........... int right_node = 2 * node + 2; int mid = (start + end) / 2; if (idx = start) { update(arr, tree, left_node, start, mid, idx, val); } else if (idx = mid + 1) { update(arr, tree, right_node, mid + 1, end, idx, val); } tree[node] = tree[left_node] + tree[right_node]; // 核心代码 } } // querey public static int querey(int tree[], int arr[], int node, int start, int end, int L, int R) { if (R < start || end < L) { // 排除范围之外 return 0; } else if (L <= start && end <= R) return tree[node]; else if (start == end) return node; else { int mid = (start + end) / 2; int left_node = node * 2 + 1; int right_node = node * 2 + 2; int sum_left = querey(tree, arr, left_node, start, mid, L, R); int sum_right = querey(tree, arr, right_node, mid + 1, end, L, R); return sum_left + sum_right; } } }