segment trees implementation

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from math import  log2, ceil
#st is segment tree
#if no of nodes is perfect squre then size of seg tree array=2N-1
#if no of nodes is not perfect sqaure then we should take nearest greater perfect square (lets say x). then segment tree size=2x-1
def make(arr):
    n=len(arr)
    h=ceil(log2(n))
    size=2*2**h-1
    st=[None]*size
    print(arr)
    #l--->left pointer
    #r---->right pointer
    #i----->index to store the sum in seg tree array
    #child 1--> stores at 2*i+1
    #child 2 stores at 2*i+2
    def build(i,l,r):
        if l>r:
            return
        elif l==r:#like merge sort base condtions
            st[i]=arr[l]
        else:
            mid=(l+r)//2
            build(2*i+1,l,mid)
            build(2*i+2,mid+1,r)
            st[i]=st[2*i+1]+st[2*i+2]
    #l,r---->constants
    # we have to compute the sum of elements from index l to r including them
    #so these are fixed
    #s---->left pointer
    #r---->right pointer
    def findsum(i,l,r,s,e):
        #complete over lap
        if l<=s and e<=r:
            return st[i]
        #no overlap
        elif e<l or s>r:
            return 0
            #partial overlap
        else:
            mid=(s+e)//2
            one=findsum(2*i+1,l,r,s,mid)
            two=findsum(2*i+2,l,r,mid+1,e)
            return one+two
    build(0,0,n-1)
    print(st)#print segment tree
    print(findsum(0,1,3,0,n-1))
    #now we shall perform updates on segment tree --->O(log n)
    ci=1#index where we have to change value
    cv=2#value stores at the index that we have to change
    diff=arr[ci]-cv#difference that we have to updated in segment tree
    #make change in array first
    arr[ci]=cv
    def update(i,start,end):
        #if the node at index i holds the cv variable in its range.Lets substract
        if start<=ci<=end:
            #some leaves might be None if the size(arr) is not perfect square
            #lets donnt disturb it
            if st[i]==None:
                return
            st[i]-=diff
            mid=(start+end)//2
            if start==end:
                return
            update(2*i+1,start,mid)
            update(2*i+2,mid+1,end)
    #update functon
    update(0,0,n-1)
    print(findsum(0,1,3,0,n-1))
    print(st)
make([1,3,5,7,9,11])