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
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);
}
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);
}
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
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
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。
总结
线段树可以很好地处理区间问题，如区间求和，求最大最小值等。
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