Hello Everyone,
Let us consider the following problem to understand Segment Trees without recursion.
We have an array arr[0 . . . n1]. We should be able to,
 Find the sum of elements from index l to r where 0 <= l <= r <= n1
 Change the value of a specified element of the array to a new value x. We need to do arr[i] = x where 0 <= i <= n1.
A simple solution is to run a loop from l to r and calculate the sum of elements in the given range. To update a value, simply do arr[i] = x. The first operation takes O(n) time and the second operation takes O(1) time.
Another solution is to create another array and store the sum from start to i at the ith index in this array. The sum of a given range can now be calculated in O(1) time, but the update operation takes O(n) time now. This works well if the number of query operations is large and there are very few updates.
What if the number of queries and updates are equal? Can we perform both the operations in O(log n) time once given the array? We can use a Segment Tree to do both operations in O(Logn) time. In this post, we will discuss the easier and yet efficient implementation of segment trees than in the previous post.
We will implement all of these multiplication and addition operations using bitwise operators.
Let us have a look at the complete implementation:
#include <bits/stdc++.h>
using
namespace
std;
// limit for array size
const
int
N = 100000;
int
n;
// array size
// Max size of tree
int
tree[2 * N];
// function to build the tree
void
build(
int
arr[])
{
// insert leaf nodes in tree
for
(
int
i=0; i<n; i++)
tree[n+i] = arr[i];
// build the tree by calculating parents
for
(
int
i = n  1; i > 0; i)
tree[i] = tree[i<<1] + tree[i<<1  1];
}
// function to update a tree node
void
updateTreeNode(
int
p,
int
value)
{
// set value at position p
tree[p+n] = value;
p = p+n;
// move upward and update parents
for
(
int
i=p; i > 1; i >>= 1)
tree[i>>1] = tree[i] + tree[i^1];
}
// function to get sum on interval [l, r)
int
query(
int
l,
int
r)
{
int
res = 0;
// loop to find the sum in the range
for
(l += n, r += n; l < r; l >>= 1, r >>= 1)
{
if
(l&1)
res += tree[l++];
if
(r&1)
res += tree[r];
}
return
res;
}
// driver program to test the above function
int
main()
{
int
a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
// n is global
n =
sizeof
(a)/
sizeof
(a[0]);
// build tree
build(a);
// print the sum in range(1,2) indexbased
cout << query(1, 3)<<endl;
// modify element at 2nd index
updateTreeNode(2, 1);
// print the sum in range(1,2) indexbased
cout << query(1, 3)<<endl;
return
0;
}
Output:
5 3
Yes! That is all. The complete implementation of the segment tree includes the query and update functions in a lower number of lines of code than the previous recursive one. Let us now understand how each of the functions works:

The picture makes it clear that the leaf nodes are stored at i+n, so we can clearly insert all leaf nodes directly.

The next step is to build the tree and it takes O(n) time. The parent always has its less index than its children, so we just process all the nodes in decreasing order, calculating the value of the parent node. If the code inside the build function to calculate parents seems confusing, then you can see this code. It is equivalent to that inside the build function.
tree[i]=tree[2i]+tree[2i+1]

Updating a value at any position is also simple and the time taken will be proportional to the height of the tree. We only update values in the parents of the given node which is being changed. So to get the parent, we just go up to the parent node, which is p/2 or p>>1, for node p. p^1 turns (2i) to (2i + 1) and vice versa to get the second child of p.

Computing the sum also works in O(log(n)) time. If we work through an interval of [3,11), we need to calculate only for nodes 19,26,12, and 5 in that order.
The idea behind the query function is whether we should include an element in the sum or whether we should include its parent. Letâ€™s look at the image once again for proper understanding. Consider that L is the left border of an interval and R is the right border of the interval [L,R). It is clear from the image that if L is odd, then it means that it is the right child of its parent and our interval includes only L and not the parent. So we will simply include this node to sum and move to the parent of its next node by doing L = (L+1)/2. Now, if L is even, then it is the left child of its parent and the interval includes its parent also unless the right borders interfere. Similar conditions are applied to the right border also for faster computation. We will stop this iteration once the left and right borders meet.
The theoretical time complexities of both previous implementation and this implementation is the same, but practically, it is found to be much more efficient as there are no recursive calls. We simply iterate over the elements that we need. Also, this is very easy to implement.
Time Complexities:
 Tree Construction: O( n )
 Query in Range: O( Log n )
 Updating an element: O( Log n ).