Persistent Segment Tree

Hello Everyone,

Consider the segment tree with green nodes . Lets call this segment tree as version-0 . The left child for each node is connected with solid red edge where as the right child for each node is connected with solid purple edge. Clearly, this segment tree consists of 15 nodes.

Now consider we need to make change in the leaf node 13 of version-0.
So, the affected nodes will be – node 13 , node 6 , node 3 , node 1 .
Therefore, for the new version (Version-1) we need to create only these 4 new nodes .

Now, lets construct version-1 for this change in segment tree. We need a new node 1 as it is affected by change done in node 13. So , we will first create a new node 1′ (yellow color) . The left child for node 1′ will be the same for left child for node 1 in version-0. So, we connect the left child of node 1′ with node 2 of version-0(red dashed line in figure). Let’s now examine the right child for node 1′ in version-1. We need to create a new node as it is affected . So we create a new node called node 3′ and make it the right child for node 1′(solid purple edge connection).

In the similar fashion we will now examine for node 3′ . The left child is affected , So we create a new node called node 6′ and connect it with solid red edge with node 3′ , where as the right child for node 3′ will be the same as right child of node 3 in version-0. So, we will make the right child of node 3 in version-0 as the right child of node 3′ in version-1(see the purple dash edge.)

Same procedure is done for node 6′ and we see that the left child of node 6′ will be the left child of node 6 in version-0(red dashed connection) and right child is newly created node called node 13′ (solid purple dashed edge).
Each yellow color node is a newly created node and dashed edges are the inter-connection between the different versions of the segment tree.

Now, the Question arises : How to keep track of all the versions?
– We only need to keep track the first root node for all the versions and this will serve the purpose to track all the newly created nodes in the different versions. For this purpose we can maintain an array of pointers to the first node of segment trees for all versions.

Let’s consider a very basic problem to see how to implement persistence in segment tree

Problem : Given an array A[] and different point update operations.Considering each point operation to create a new version of the array. We need to answer the queries of type Q v l r : output the sum of elements in range l to r just after the v-th update.

We will create all the versions of the segment tree and keep track of their root node.Then for each range sum query we will pass the required version’s root node in our query function and output the required sum.

Below is the implementation for the above problem:-

// C++ program to implement persistent segment

// tree.

#include "bits/stdc++.h"

using namespace std;

#define MAXN 100

/* data type for individual

* node in the segment tree */

struct node


// stores sum of the elements in node

int val;

// pointer to left and right children

node* left, *right;

// required constructors........

node() {}

node(node* l, node* r, int v)


left = l;

right = r;

val = v;



// input array

int arr[MAXN];

// root pointers for all versions

node* version[MAXN];

// Constructs Version-0

// Time Complexity : O(nlogn)

void build(node* n, int low, int high)


if (low==high)


n->val = arr[low];

return ;


int mid = (low+high) / 2;

n->left = new node(NULL, NULL, 0);

n->right = new node(NULL, NULL, 0);

build(n->left, low, mid);

build(n->right, mid+1, high);

n->val = n->left->val + n->right->val;



* Upgrades to new Version

* @param prev : points to node of previous version

* @param cur : points to node of current version

* Time Complexity : O(logn)

* Space Complexity : O(logn) */

void upgrade(node* prev, node* cur, int low, int high,

int idx, int value)


if (idx > high or idx < low or low > high)

return ;

if (low == high)


// modification in new version

cur->val = value;

return ;


int mid = (low+high) / 2;

if (idx <= mid)


// link to right child of previous version

cur->right = prev->right;

// create new node in current version

cur->left = new node(NULL, NULL, 0);

upgrade(prev->left,cur->left, low, mid, idx, value);




// link to left child of previous version

cur->left = prev->left;

// create new node for current version

cur->right = new node(NULL, NULL, 0);

upgrade(prev->right, cur->right, mid+1, high, idx, value);


// calculating data for current version

// by combining previous version and current

// modification

cur->val = cur->left->val + cur->right->val;


int query(node* n, int low, int high, int l, int r)


if (l > high or r < low or low > high)

return 0;

if (l <= low and high <= r)

return n->val;

int mid = (low+high) / 2;

int p1 = query(n->left,low,mid,l,r);

int p2 = query(n->right,mid+1,high,l,r);

return p1+p2;


int main( int argc, char const *argv[])


int A[] = {1,2,3,4,5};

int n = sizeof (A)/ sizeof ( int );

for ( int i=0; i<n; i++)

arr[i] = A[i];

// creating Version-0

node* root = new node(NULL, NULL, 0);

build(root, 0, n-1);

// storing root node for version-0

version[0] = root;

// upgrading to version-1

version[1] = new node(NULL, NULL, 0);

upgrade(version[0], version[1], 0, n-1, 4, 1);

// upgrading to version-2

version[2] = new node(NULL, NULL, 0);

upgrade(version[1],version[2], 0, n-1, 2, 10);

cout << "In version 1 , query(0,4) : " ;

cout << query(version[1], 0, n-1, 0, 4) << endl;

cout << "In version 2 , query(3,4) : " ;

cout << query(version[2], 0, n-1, 3, 4) << endl;

cout << "In version 0 , query(0,3) : " ;

cout << query(version[0], 0, n-1, 0, 3) << endl;

return 0;



In version 1 , query(0,4) : 11 In version 2 , query(3,4) : 5 In version 0 , query(0,3) : 10

Time Complexity : The time complexity will be the same as the query and point update operation in the segment tree as we can consider the extra node creation step to be done in O(1). Hence, the overall Time Complexity per query for new version creation and range sum query will be O(log n) .