Java数据结构之线段树详解

介绍

线段树(又名区间树)也是一种二叉树，每个节点的值等于左右孩子节点值的和，线段树示例图如下

以求和为例，根节点表示区间0-5的和，左孩子表示区间0-2的和，右孩子表示区间3-5的和，依次类推。

代码实现

/**
* 使用数组实现线段树
*/
public class SegmentTree<E> {

private Node[] data;
 private int size;

private Merger<E> merger;

public SegmentTree(E[] source, Merger<E> merger) {
   this.merger = merger;
   this.size = source.length;
   this.data = new Node[size * 4];
   buildTree(0, source, 0, size - 1);
 }

public E search(int queryLeft, int queryRight) {
   if (queryLeft < 0 || queryLeft > size || queryRight < 0 || queryRight > size
       || queryLeft > queryRight) {
     throw new IllegalArgumentException("index is illegal");
   }
   return search(0, queryLeft, queryRight);
 }

/**
  * 查询区间queryLeft-queryRight的值
  */
 private E search(int treeIndex, int queryLeft, int queryRight) {
   Node treeNode = data[treeIndex];
   int left = treeNode.left;
   int right = treeNode.right;
   if (left == queryLeft && right == queryRight) {
     return elementData(treeIndex);
   }
   int leftTreeIndex = leftChild(treeIndex);
   int rightTreeIndex = rightChild(treeIndex);
   int middle = left + ((right - left) >> 1);
   if (queryLeft > middle) {
     return search(rightTreeIndex, queryLeft, queryRight);
   } else if (queryRight <= middle) {
     return search(leftTreeIndex, queryLeft, queryRight);
   }
   E leftEle = search(leftTreeIndex, queryLeft, middle);
   E rightEle = search(rightTreeIndex, middle + 1, queryRight);
   return merger.merge(leftEle, rightEle);
 }

public void update(int index, E e) {
   update(0, index, e);
 }

/**
  * 更新索引为index的值为e
  */
 private void update(int treeIndex, int index, E e) {
   Node treeNode = data[treeIndex];
   int left = treeNode.left;
   int right = treeNode.right;
   if (left == right) {
     treeNode.data = e;
     return;
   }
   int leftTreeIndex = leftChild(treeIndex);
   int rightTreeIndex = rightChild(treeIndex);
   int middle = left + ((right - left) >> 1);
   if (index > middle) {
     update(rightTreeIndex, index, e);
   } else {
     update(leftTreeIndex, index, e);
   }
   treeNode.data = merger.merge(elementData(leftTreeIndex), elementData(rightTreeIndex));
 }

private void buildTree(int treeIndex, E[] source, int left, int right) {
   if (left == right) {
     data[treeIndex] = new Node<>(source[left], left, right);
     return;
   }
   int leftTreeIndex = leftChild(treeIndex);
   int rightTreeIndex = rightChild(treeIndex);
   int middle = left + ((right - left) >> 1);
   buildTree(leftTreeIndex, source, left, middle);
   buildTree(rightTreeIndex, source, middle + 1, right);
   E treeData = merger.merge(elementData(leftTreeIndex), elementData(rightTreeIndex));
   data[treeIndex] = new Node<>(treeData, left, right);
 }

@Override
 public String toString() {
   return Arrays.toString(data);
 }

private E elementData(int index) {
   return (E) data[index].data;
 }

private int leftChild(int index) {
   return index * 2 + 1;
 }

private int rightChild(int index) {
   return index * 2 + 2;
 }

private static class Node<E> {

E data;
   int left;
   int right;

Node(E data, int left, int right) {
     this.data = data;
     this.left = left;
     this.right = right;
   }

@Override
   public String toString() {
     return String.valueOf(data);
   }
 }

public interface Merger<E> {

E merge(E e1, E e2);
 }
}

我们以LeetCode上的一个问题来分析线段树的构建，查询和更新，LeetCode307问题如下：

给定一个整数数组，查询索引区间[i,j]的元素的总和。

线段树构建

private void buildTree(int treeIndex, E[] source, int left, int right) {
   if (left == right) {
     data[treeIndex] = new Node<>(source[left], left, right);
     return;
   }
   int leftTreeIndex = leftChild(treeIndex);
   int rightTreeIndex = rightChild(treeIndex);
   int middle = left + ((right - left) >> 1);
   buildTree(leftTreeIndex, source, left, middle);
   buildTree(rightTreeIndex, source, middle + 1, right);
   E treeData = merger.merge(elementData(leftTreeIndex), elementData(rightTreeIndex));
   data[treeIndex] = new Node<>(treeData, left, right);
 }

测试代码

public class Main {

public static void main(String[] args) {
   Integer[] nums = {-2, 0, 3, -5, 2, -1};
   SegmentTree<Integer> segmentTree = new SegmentTree<>(nums, Integer::sum);
   System.out.println(segmentTree);
 }

}

最后构造出的线段树如下，前面为元素值，括号中为包含的区间。

递归构造过程为

• 当左指针和右指针相等时，表示为叶子节点

• 将左孩子和右孩子值相加，构造当前节点，依次类推

区间查询

/**
  * 查询区间queryLeft-queryRight的值
  */
 private E search(int treeIndex, int queryLeft, int queryRight) {
   Node treeNode = data[treeIndex];
   int left = treeNode.left;
   int right = treeNode.right;
   if (left == queryLeft && right == queryRight) {
     return elementData(treeIndex);
   }
   int leftTreeIndex = leftChild(treeIndex);
   int rightTreeIndex = rightChild(treeIndex);
   int middle = left + ((right - left) >> 1);
   if (queryLeft > middle) {
     return search(rightTreeIndex, queryLeft, queryRight);
   } else if (queryRight <= middle) {
     return search(leftTreeIndex, queryLeft, queryRight);
   }
   E leftEle = search(leftTreeIndex, queryLeft, middle);
   E rightEle = search(rightTreeIndex, middle + 1, queryRight);
   return merger.merge(leftEle, rightEle);
 }

查询区间2-5的和

public class Main {

public static void main(String[] args) {
   Integer[] nums = {-2, 0, 3, -5, 2, -1};
   SegmentTree<Integer> segmentTree = new SegmentTree<>(nums, Integer::sum);
   System.out.println(segmentTree);
   System.out.println(segmentTree.search(2, 5)); // -1
 }

}

查询过程为

• 待查询的区间和当前节点的区间相等，返回当前节点值

• 待查询左区间大于中间区间值，查询右孩子

• 待查询右区间小于中间区间值，查询左孩子

• 待查询左区间在左孩子，右区间在右孩子，两边查询结果相加

更新

/**
  * 更新索引为index的值为e
  */
 private void update(int treeIndex, int index, E e) {
   Node treeNode = data[treeIndex];
   int left = treeNode.left;
   int right = treeNode.right;
   if (left == right) {
     treeNode.data = e;
     return;
   }
   int leftTreeIndex = leftChild(treeIndex);
   int rightTreeIndex = rightChild(treeIndex);
   int middle = left + ((right - left) >> 1);
   if (index > middle) {
     update(rightTreeIndex, index, e);
   } else {
     update(leftTreeIndex, index, e);
   }
   treeNode.data = merger.merge(elementData(leftTreeIndex), elementData(rightTreeIndex));
 }

更新只影响元素值，不影响元素区间。

更新其实和构建的逻辑类似，找到待更新的实际索引，依次更新父节点的值。

来源：https://www.cnblogs.com/strongmore/p/14223224.html