how to count no of left or right child from binary tree database

Question

This is a dummy of my table

• first node is 1
• LorR define child is ini left(0) or Right(1)
• 1 st node has two child node 2 and 3 in Left and Right
• 2 nd node has two child node 4 and 5 in Left and Right
• 3 rd node has two child node 6 and 7 in Left and Right
• 4 th node has two child node 8 and 9 in Left and Right
• 5 th node has one child node 10 in Left
• 6 th node has one child node 11 in Left
• 7 th node has one child node 12 in Right

now if i want to find the number of total child nodes of any parent node in each level for ex...

• node 2 (parent) has a 2 child (4,5) in level 1 & 2 child node (8,9) in level 2
• node 6 (parent) has a 1 child (11) in level 1
• node 3 (parent) has a 2 child (6,7) in level 1 & 2 child nodes (11,12) in level 2

how can I implement this using php? My code is code

Answer 1

Based on what I am understanding, you should transform the table into informative PHP array, something like:

$a = array(
    1 => array(
        'children' => array(
            array(2, 'left'),
            array(3, 'right'),
        ),
        'level' => 0
    ),
    2 => array(
        'children' => array(
            array(4, 'left'),
            array(5, 'right'),
        ),
        'level' => 1
    ),
...        
    12 => array(
        'children' => array(),
        'level' => 3
    ),
);

This task is quite easy, just fetch the table then foreach and assign each row into associated array. Here is the function to get all children count:

function getChildrenCount($nodeId, $a, $rootLevel)
{
    $leftChildren = array();
    $rightChildren = array();
    $countCurrentNode = 0;
    $level = $a[$nodeId]['level'] - $rootLevel;
    if (empty($a[$nodeId]['children'])) {
        return array($level => 1);
    } else {
        foreach ($a[$nodeId]['children'] as $children) {
            if ($children[1] == 'left') {
                $leftChildren = getChildrenCount($children[0], $a, $rootLevel);
                $countCurrentNode++;
            }
            if ($children[1] == 'right') {
                $rightChildren = getChildrenCount($children[0], $a, $rootLevel);
                $countCurrentNode++;
            }
        }

        $current = $leftChildren;
        foreach ($rightChildren as $rightLevel => $count) {
            if (isset($current[$rightLevel])) {
                $current[$rightLevel] += $count;
            } else {
                $current[$rightLevel] = $count;
            }
        }
        $current[$level] = $countCurrentNode;
        return $current;
    }
}

The idea is the recursive traversal each node, then calculate number of children based on its level (compared with root level)

How to call:

getChildrenCount(2, $a, $a[2]['level'])

It will return array(level => count):

array (size=3)
0 => int 2 // level 1
1 => int 3 // level 2
2 => int 3 // Should be removed

Then you should remove the last element - it's useless as it's used to count child by level.

Note:

1. It should have a better way to organize the data (array $a), to have easier getChildrenCount() implementation.
2. This code is not tested carefully, just checked with some cases.