A Segment Tree is a data structure that can be used to perform range queries and range updates. It is a height-balanced binary tree, usually built on top of an Array. Segment Trees can be used to solve Range Min/Max & Sum Queries and Range Update Queries in O(log n) time.
The Segment Tree works like other tree data structures. It creates query paths that limit the amount of processing required to return data. Each intermediate node of the tree represents a segment of the data set. The root node contains the sum of all numbers in the tree. Its children contain the sums of all the numbers in their respective ranges, and so on down the tree to leaf nodes.
Sum all elements in a range.
Find the min or max value of elements in a range.
Update all elements in a range.
This doesn’t mean that using Segment Trees is limited to working with numbers. You can work with Segment Trees, for example, to find all the intervals (or ranges) that match a specific criteria. The classic example of this is the Brackets Problem.
// If segment of this node is a part of given range, then
// return the min of the segment
if(qs <= ss && qe >= se)
returnst[index];
// If segment of this node is outside the given range
if(se < qs || ss > qe)
returnInteger.MAX_VALUE;
// If a part of this segment overlaps with the given range
intmid = getMid(ss, se);
returnminVal(RMQUtil(ss, mid, qs, qe, 2* index + 1),
RMQUtil(mid + 1, se, qs, qe, 2* index + 2));
}
// Return minimum of elements in range from index qs (quey
// start) to qe (query end). It mainly uses RMQUtil()
intRMQ(intn, intqs, intqe)
{
// Check for erroneous input values
if(qs < 0|| qe > n - 1|| qs > qe) {
System.out.println("Invalid Input");
return-1;
}
returnRMQUtil(0, n - 1, qs, qe, 0);
}
// A recursive function that constructs Segment Tree for
// array[ss..se]. si is index of current node in segment tree st
intconstructSTUtil(intarr[], intss, intse, intsi)
{
// If there is one element in array, store it in current
// node of segment tree and return
if(ss == se) {
st[si] = arr[ss];
returnarr[ss];
}
// If there are more than one elements, then recur for left and
// right subtrees and store the minimum of two values in this node
intmid = getMid(ss, se);
st[si] = minVal(constructSTUtil(arr, ss, mid, si * 2+ 1),
constructSTUtil(arr, mid + 1, se, si * 2+ 2));
returnst[si];
}
https://www.quora.com/What-is-a-segment-tree-and-what-are-its-applications Segment tree is a data structure which can be used to perform range queries and range updates. The underlying principle behind this data structure is to store certain values of ranges as a balanced binary tree and hence query and update can be performed efficiently.
You are given an array of 'n' elements. You want to perform the following operations on the array. 1) Choose a segment [l,r] and increment all the elements in that range by 1. 2) Given a segment [l,r], obtain the sum of elements in that range.
The segment tree is a highly versatile data structure, based upon the divide-and-conquer paradigm, which can be thought of as a tree of intervals of an underlying array, constructed so that queries on ranges of the array as well as modifications to the array's elements may be efficiently performed.
The divide-and-conquer solution
The divide-and-conquer solution would be as follows:
If the range contains one element, that element itself is trivially the minimum within that range.
Otherwise, divide the range into two smaller ranges, each approximately half the size of the original, and find their respective minima. The minimum for the original range is then the smaller of the two minima of the sub-ranges
1. Leaf Nodes are the elements of the input array. 2. Each internal node represents some merging of the leaf nodes. The merging may be different for different problems. For this problem, merging is sum of leaves under a node.
An array representation of tree is used to represent Segment Trees. For each node at index i, the left child is at index 2*i+1, right child at 2*i+2 and the parent is at .
Construction of Segment Tree from given array
We start with a segment arr[0 . . . n-1]. and every time we divide the current segment into two halves(if it has not yet become a segment of length 1), and then call the same procedure on both halves, and for each such segment we store the sum in corresponding node.
All levels of the constructed segment tree will be completely filled except the last level. Also, the tree will be a Full Binary Tree because we always divide segments in two halves at every level. Since the constructed tree is always full binary tree with n leaves, there will be n-1 internal nodes. So total number of nodes will be 2*n – 1.
Height of the segment tree will be . Since the tree is represented using array and relation between parent and child indexes must be maintained, size of memory allocated for segment tree will be .
classSegmentTree
{
intst[]; // The array that stores segment tree nodes
/* Constructor to construct segment tree from given array. This
constructor allocates memory for segment tree and calls
publicclass SegmentTree {
privatestaticint[] st;
//Construction of Segment Treepublicstaticint[] constructSegmentTree(int[] arr){
int height = (int)Math.ceil(Math.log(arr.length)/Math.log(2));
int size = 2*(int)Math.pow(2, height)-1;
st = newint[size];
constructST(arr, 0, arr.length-1, 0);
return st;
}
publicstaticint constructST(int[] arr, int ss, int se, int si){
if(ss==se){
st[si] = arr[ss];
return st[si];
}
int mid = ss+(se-ss)/2;
st[si] = constructST(arr, ss, mid, si*2+1)+
constructST(arr, mid+1, se, si*2+2);
return st[si];
}
//Finding The sum of given Rangepublicstaticint findRangeSum(int ss, int se, int[] arr){
int n = arr.length;
if(ss<0 || se > n-1 || ss>se){
thrownewIllegalArgumentException("Invalid arguments");
}
return findSum(0, n-1, ss, se, 0);
}
publicstaticint findSum(int ss, int se, int qs, int qe, int si){
if(ss>qe || se < qs)return 0;
if(qs<=ss && qe>=se)return st[si];
int mid = ss+(se-ss)/2;
return findSum(ss, mid, qs, qe, si*2+1)+
findSum(mid+1, se, qs, qe, si*2+2);
}
//Updating a particular index's valuestaticvoid updateValue(int arr[], int i, int newVal){
if(i<0 || i>arr.length-1)thrownewIllegalArgumentException();
int difference = newVal - arr[i];
arr[i] = newVal;
updValue(arr, 0, arr.length-1, i, difference, 0);
}
staticvoid updValue(int[] arr, int ss, int se, int i, int difference, int si){
if(i< ss || i>se)return;
st[si] = st[si]+difference;
if(ss!=se){
int mid = ss+(se-ss)/2;
updValue(arr, ss, mid, i, difference, si*2+1);
updValue(arr, mid+1, se, i, difference, si*2+2);
}
}
}
* The <tt>SegmentTree</tt> class is an structure for efficient search of cummulative data. * It performs Range Minimum Query and Range Sum Query in O(log(n)) time. * It can be easily customizable to support Range Max Query, Range Multiplication Query etc. * <p/> * Also it has been develop with <tt>LazyPropagation</tt> for range updates, which means * when you perform update operations over a range, the update process affects the least nodes as possible * so that the bigger the range you want to update the less time it consumes to update it. Eventually those changes will be propagated * to the children and the whole array will be up to date.
publicclassSegmentTree{private Node[] heap;privateint[] array;privateint size;/** * Time-Complexity: O(n*log(n)) * * @param array the Initialization array */publicSegmentTree(int[] array){this.array = Arrays.copyOf(array, array.length);//The max size of this array is about 2 * 2 ^ log2(n) + 1 size =(int)(2* Math.pow(2.0, Math.floor((Math.log((double) array.length)/ Math.log(2.0))+1))); heap =new Node[size];build(1,0, array.length);}publicintsize(){return array.length;}//Initialize the Nodes of the Segment treeprivatevoidbuild(int v,int from,int size){ heap[v]=newNode(); heap[v].from = from; heap[v].to = from + size -1;if(size ==1){ heap[v].sum = array[from]; heap[v].min = array[from];}else{//Build childsbuild(2* v, from, size /2);build(2* v +1, from + size /2, size - size /2); heap[v].sum = heap[2* v].sum + heap[2* v +1].sum;//min = min of the children heap[v].min = Math.min(heap[2* v].min, heap[2* v +1].min);}}/** * Range Sum Query * * Time-Complexity: O(log(n)) */publicintRSQ(int from,int to){returnRSQ(1, from, to);}privateintRSQ(int v,int from,int to){Node n = heap[v];//If you did a range update that contained this node, you can infer the Sum without going down the treeif(n.pendingVal !=null&&contains(n.from, n.to, from, to)){return(to - from +1)* n.pendingVal;}if(contains(from, to, n.from, n.to)){return heap[v].sum;}if(intersects(from, to, n.from, n.to)){propagate(v);int leftSum =RSQ(2* v, from, to);int rightSum =RSQ(2* v +1, from, to);return leftSum + rightSum;}return0;}/** * Range Min Query * * Time-Complexity: O(log(n)) */publicintRMinQ(int from,int to){returnRMinQ(1, from, to);}privateintRMinQ(int v,int from,int to){Node n = heap[v];//If you did a range update that contained this node, you can infer the Min value without going down the treeif(n.pendingVal !=null&&contains(n.from, n.to, from, to)){return n.pendingVal;}if(contains(from, to, n.from, n.to)){return heap[v].min;}if(intersects(from, to, n.from, n.to)){propagate(v);int leftMin =RMinQ(2* v, from, to);int rightMin =RMinQ(2* v +1, from, to);return Math.min(leftMin, rightMin);}return Integer.MAX_VALUE;}/** * Range Update Operation. * With this operation you can update either one position or a range of positions with a given number. * The update operations will update the less it can to update the whole range (Lazy Propagation). * The values will be propagated lazily from top to bottom of the segment tree. * This behavior is really useful for updates on portions of the array * <p/> * Time-Complexity: O(log(n)) * * @param from * @param to * @param value */publicvoidupdate(int from,int to,int value){update(1, from, to, value);}privatevoidupdate(int v,int from,int to,int value){//The Node of the heap tree represents a range of the array with bounds: [n.from, n.to]Node n = heap[v];/** * If the updating-range contains the portion of the current Node We lazily update it. * This means We do NOT update each position of the vector, but update only some temporal * values into the Node; such values into the Node will be propagated down to its children only when they need to. */if(contains(from, to, n.from, n.to)){change(n, value);}if(n.size()==1)return;if(intersects(from, to, n.from, n.to)){/** * Before keeping going down to the tree We need to propagate the * the values that have been temporally/lazily saved into this Node to its children * So that when We visit them the values are properly updated */propagate(v);update(2* v, from, to, value);update(2* v +1, from, to, value); n.sum = heap[2* v].sum + heap[2* v +1].sum; n.min = Math.min(heap[2* v].min, heap[2* v +1].min);}}//Propagate temporal values to childrenprivatevoidpropagate(int v){Node n = heap[v];if(n.pendingVal !=null){change(heap[2* v], n.pendingVal);change(heap[2* v +1], n.pendingVal); n.pendingVal =null;//unset the pending propagation value}}//Save the temporal values that will be propagated lazilyprivatevoidchange(Node n,int value){ n.pendingVal = value; n.sum = n.size()* value; n.min = value; array[n.from]= value;}//Test if the range1 contains range2privatebooleancontains(int from1,int to1,int from2,int to2){return from2 >= from1 && to2 <= to1;}//check inclusive intersection, test if range1[from1, to1] intersects range2[from2, to2]privatebooleanintersects(int from1,int to1,int from2,int to2){return from1 <= from2 && to1 >= from2 // (.[..)..] or (.[...]..)|| from1 >= from2 && from1 <= to2;// [.(..]..) or [..(..)..}//The Node class represents a partition range of the array.staticclassNode{int sum;int min;//Here We store the value that will be propagated lazilyInteger pendingVal =null;int from;int to;intsize(){return to - from +1;}}
The problem that a segment tree can solve is the following. We are given an array of values a[0], a[1], ..., a[N - 1]. Assume without loss of generality that N = 2n; we can generally pad the computations accordingly. Also, consider some associative binary function f. Examples of f include sum, min, max, or gcd (as in the Timus problem). Segment trees allow for each of the following two operations onO(logN) time:
compute f(a[i], a[i + 1], ..., a[j]) for i ≤ j; and
public class SegmentTree
{
private int[] tree;
private int maxsize;
private int height;
private final int STARTINDEX = 0;
private final int ENDINDEX;
private final int ROOT = 0;
public SegmentTree(int size)
{
height = (int)(Math.ceil(Math.log(size) / Math.log(2)));
maxsize = 2 * (int) Math.pow(2, height) - 1;
tree = new int[maxsize];
ENDINDEX = size - 1;
}
private int leftchild(int pos)
{
return 2 * pos + 1;
}
private int rightchild(int pos)
{
return 2 * pos + 2;
}
private int mid(int start, int end)
{
return (start + (end - start) / 2);
}
private int getSumUtil(int startIndex, int endIndex, int queryStart, int queryEnd, int current)
{
if (queryStart <= startIndex && queryEnd >= endIndex )
{
return tree[current];
}
if (endIndex < queryStart || startIndex > queryEnd)
{
return 0;
}
int mid = mid(startIndex, endIndex);
return getSumUtil(startIndex, mid, queryStart, queryEnd, leftchild(current))
+ getSumUtil( mid + 1, endIndex, queryStart, queryEnd, rightchild(current));
}
public int getSum(int queryStart, int queryEnd)
{
if(queryStart < 0 || queryEnd > tree.length)
{
return -1;
}
return getSumUtil(STARTINDEX, ENDINDEX, queryStart, queryEnd, ROOT);
}
private int constructSegmentTreeUtil(int[] elements, int startIndex, int endIndex, int current)
{
if (startIndex == endIndex)
{
tree[current] = elements[startIndex];
return tree[current];
}
int mid = mid(startIndex, endIndex);
tree[current] = constructSegmentTreeUtil(elements, startIndex, mid, leftchild(current))
+ constructSegmentTreeUtil(elements, mid + 1, endIndex, rightchild(current));
return tree[current];
}
public void constructSegmentTree(int[] elements)
{
constructSegmentTreeUtil(elements, STARTINDEX, ENDINDEX, ROOT);
}
private void updateTreeUtil(int startIndex, int endIndex, int updatePos, int update, int current)
{
if ( updatePos < startIndex || updatePos > endIndex)
{
return;
}
tree[current] = tree[current] + update;
if (startIndex != endIndex)
{
int mid = mid(startIndex, endIndex);
updateTreeUtil(startIndex, mid, updatePos, update, leftchild(current));
updateTreeUtil(mid+1, endIndex, updatePos, update, rightchild(current));
}
}
public void update(int update, int updatePos, int[] elements)
{
int updatediff = update - elements[updatePos] ;
elements[updatePos] = update;
updateTreeUtil(STARTINDEX, ENDINDEX, updatePos, updatediff, ROOT);
}
}
http://blueocean-penn.blogspot.com/2015/01/segment-tree-order-statistics-binary.html Segment tree is a binary tree which stores all original elements are leaves in sequential order of their indices, and internal nodes of the tree store the aggregate information of their children. The tree is often represented as binary heap-like data structure. Node i's left children is 2*i+1, right children is 2*i+2. The height of the tree is Ceiling(lg(SIZE_OF_INPUT)). For example, array of 8 elements (0, 1, ..., 7), its height will be 3, array of 6 elements, its height will be 3 too. So the heap storage size will be (1 + 2 + 4 + .. + 2^height) = 2^(height+1)-1. Note the size of the tree is 2*SIZE_OF_INPUT-1.
The segment tree is a static data structure, that is, in order to construct the tree the original input has to be known first. However it can be used to solve on-line or streaming problem. For example, we can initialized the tree to a large size with values for any future incoming elements are set to default values. For example, for min query, we can set the default values to Integer.MIN_VALUE. In this way, any insertion becomes update operation. Of course the drawback of this implementation is that we are wasting some space in the tree implementation.
* a segment tree representation using array
* note this supports min query only, but can be adopted to other types like sum, max and etc.
*/
publicclass SegmentTree {
int[] tree;
intlen;
public SegmentTree(int[] nums){
len = nums.length;
int x = (int) Math.ceil(Math.log10(len)/Math.log10(2));
tree = newint[(int) (Math.pow(2, x)*2-1)];
constructTree(0, nums, 0, nums.length-1);
}
//Time Complexity for tree construction is O(n). There are total 2n-1 nodes, and value of //every node is calculated only once in tree construction.
privateint constructTree(int current, int[] nums, int start, int end){
A Range Tree is also something similar except that the inserted entities are points and not ranges like in the interval tree.
A BIT (Binary Indexed Tree) is a notional tree that lets one compute range sums quickly. It's just a way to store sums on some subset of intervals that lets you construct the sum over any interval with these precomputed interval sum subsets. You can't use a BIT to compute range minimum or maximum queries since the only thing you can do with a BIT is compute the sum/min/max of a range that starts from index 0 (starting of the interval). For range sum in the range (a, b), one can compute the sum as sum(0, b) - sum(0, a), However, the same trick can't be used for min/max.
A Segment Tree is a more general notion of a BIT, where one precomputes the sum over a fixed set of ranges (not necessarily starting from index 0), which lets one layer O(log n) such ranges to compute the sum/min/max over any any arbitrary range.
.
Construction of Segment Tree from given array
. Since the tree is represented using array and relation between parent and child indexes must be maintained, size of memory allocated for segment tree will be 
.
.
. Since the tree is represented using array and relation between parent and child indexes must be maintained, size of memory allocated for segment tree will be 
.
