Segment tree and queries

Question

I studied segment tree recently and I made this code of segment tree and its queries. Is my code correct in terms of time complexity?

class segmentTree{
private:
    vector<int> mini;
    vector<int> tree;
    vector<int> maxi;
    int n;

Here in the constructor I passed the dynamic array.

public:

    segmentTree(vector<int> a)
    {
        mini=a;
        tree=a;
        maxi=a;
        n=tree.size();
    }

makeTree function builds three segment trees for different queries like sum, maximum and minimum.

    void makeTree()
    {
        reverse(tree.begin(),tree.end());
        reverse(mini.begin(),mini.end());
        reverse(maxi.begin(),maxi.end());
        for(int i=0;i<tree.size();i+=2)
        {
            tree.pb(tree[i]+tree[i+1]);
            mini.pb(min(mini[i],mini[i+1]));
            maxi.pb(max(maxi[i],maxi[i+1]));
        }
        tree.PB();
        mini.PB();
        maxi.PB();
        reverse(tree.begin(),tree.end());
        reverse(mini.begin(),mini.end());
        reverse(maxi.begin(),maxi.end());
    }

The sum function calculates the sum of given range.

    int sum(int a,int b)
    {
        a+=(n);
        b+=(n);
        int sum=0;
        while(a<=b)
        {
            if(a%2==1) 
            {
                sum+=tree[a-1];
                a++;
            }
            if(b%2==0) 
            {
                sum+=tree[b-1];
                b--;
            }
            a/=2;
            b/=2; 
        }
        return sum;
    }

The minimum function finds the minimum value of given range.

    int minimum(int a,int b)
    {
        a+=(n);
        b+=(n);
        int minu;
        while(a<=b)
        {
            if(a%2==1)
            {
                minu=mini[a-1];
                a++;
            }
            if(b%2==0)
            {
                minu=mini[b-1];
                b--;
            }
            a/=2;b/=2;
        }
        return minu;
    }

The maximum function finds the maximum value of given range.

int maximum(int a,int b)
    {
        a+=(n);
        b+=(n);
        int maxu;
        while(a<=b)
        {
            if(a%2==1)
            {
                maxu=maxi[a-1];
                a++;
            }
            if(b%2==0)
            {
                maxu=maxi[b-1];
                b--;
            }
            a/=2;b/=2;
        }
        return maxu;
    }
};