RogerBW's Blog

Perl Weekly Challenge 130: An Odd Tree 17 September 2021

I’ve been doing the Weekly Challenges. The latest involved more binary trees and linked lists. (Note that this is open until 19 September 2021.)

TASK #1 › Odd Number

You are given an array of positive integers, such that all the numbers appear even number of times except one number.

Write a script to find that integer.

One could count all the occurrences and check them for evenness, but we have hashes and sets and things, so there's no need. Here's the Raku (the Ruby and Rust are functionally identical):

sub on(@tree) {

SetHash is the mutable sort of set.

  my $;

Iterate through the list.

  for @tree -> $n {

If there's a key in the set for this number, it's now shown up an even number of times. Delete the key.

    if ($k.{$n}) {

If there was no key, the number has now shown up an odd number of times. Set the key.

    } else {

The remaining keys are therefore the numbers that have shown up an odd number of times. We're told there's exactly one of these, so return it.

  return ($k.keys)[0];

Perl and PostScript don't have sets, so I use a hash/dictionary. (Getting the hash key out is a little more fiddly in PostScript.)

/on {
    dup length dict /k exch def
        k exch known {
            k exch undef
        } {
            k exch 1 put
        } ifelse
    } forall
    k {
        /out exch def
    } forall
} def

Python does have native sets, but they're only partly mutable; you can add keys to them, but not delete them. I dug around a bit looking for documentation on MutableSet, which appears to do what I want, but ended up just using a dict instead.

TASK #2 › Binary Search Tree

You are given a tree.

Write a script to find out if the given tree is Binary Search Tree (BST).

According to wikipedia, the definition of BST:

A binary search tree is a rooted binary tree, whose internal
nodes each store a key (and optionally, an associated value),
and each has two distinguished sub-trees, commonly denoted left
and right. The tree additionally satisfies the binary search
property: the key in each node is greater than or equal to any
key stored in the left sub-tree, and less than or equal to any
key stored in the right sub-tree. The leaves (final nodes) of
the tree contain no key and have no structure to distinguish
them from one another.

Once more my preferred tree representation is helpful. I define limit, a list of tuples for each node in the tree, initially containing two copies of the value of that node. This will be updated to contain the minimum and maximum values of that node plus all child nodes. Here's the Rust:

fn bst(tree: Vec<u64>) -> i8 {
    let mut limit: Vec<(u64,u64)>=vec![];
    for i in &tree {

Iterate downwards from the last-but-one row (i.e. over all nodes that have children, branch to root).

    for s in (0..=((tree.len()-1)/2)-1).rev() {
        let child=s*2+1;
        for sb in 0..=1 {
            let ac=child+sb;

If we have a left-subtree that contains anything larger than this node's value, bail out reporting failure.

            if sb==0 && tree[s]!=0 && limit[ac].1 > tree[s] {
                return 0;

Similarly a right-subtree with anything smaller.

            if sb==1 && limit[ac].0!=0 && limit[ac].0 < tree[s] {
                return 0;

Now update the limit values for this node: the lower of our own value and the left-child's minimum, the higher of our own value and the right-child's maximum.

        if limit[child].0 > 0 {

If there were no failures, we succeeded.

    return 1;

Full code on github.

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