How to build a segment tree without recursion?

Asked Mar 28 '20 at 10:58

Active Mar 28 '20 at 13:28

Viewed 278 times

I'm trying to understand the process of building a segment tree using Python. I've come up with a function like this (which works):

arr = [ ... ]
tree = [None] * (4*len(arr))

def build(v, vl, vr):
    if vl == vr:
        tree[v] = arr[vl]
        return

    vm = (vl + vr) // 2
    build(2*v + 1, vl, vm)
    build(2*v + 2, vm + 1, vr)
    tree[v] = tree[2*v + 1] + tree[2*v + 2]

build(0, 0, len(arr) - 1)

How can I make this build function iterative (w/o recursion)? Also, what's the time complexity of building a segment tree this way? Some tutorials state that it's O(n), but don't the recursive calls make it O(n*log(n))?

edited Mar 28 '20 at 13:28

asked Mar 28 '20 at 10:58

planetp

14,248
20
86
160

If the code works, I think this is could be asked in [codereview](https://codereview.stackexchange.com/). – Dani Mesejo Mar 28 '20 at 11:01
What is `arr = [ ... ]` ? – Dani Mesejo Mar 28 '20 at 11:10
[Builds a segment Tree without recursion](https://www.geeksforgeeks.org/segment-tree-efficient-implementation/) – DarrylG Mar 28 '20 at 11:13
`arr` is the original array from which the segment tree is built. – planetp Mar 28 '20 at 13:21
It seems that you suffer from an irrational "fear" of recursion. – Mar 28 '20 at 13:31
[Efficient and easy segment trees - Codeforces](https://codeforces.com/blog/entry/18051) – martian Feb 14 '23 at 15:04

How to build a segment tree without recursion?

0 Answers0