public static KarthicAryTree SetUpAryTreeWithThreeChildren()
{
KarthicAryTree tree = new KarthicAryTree();
KarthicAryTreeNode node1 = new KarthicAryTreeNode(1);
KarthicAryTreeNode node2 = new KarthicAryTreeNode(2);
KarthicAryTreeNode node3 = new KarthicAryTreeNode(3);
KarthicAryTreeNode node4 = new KarthicAryTreeNode(4);
KarthicAryTreeNode node5 = new KarthicAryTreeNode(5);
KarthicAryTreeNode node6 = new KarthicAryTreeNode(6);
KarthicAryTreeNode node7 = new KarthicAryTreeNode(7);
KarthicAryTreeNode node8 = new KarthicAryTreeNode(8);
KarthicAryTreeNode node9 = new KarthicAryTreeNode(9);
tree.root = node1;
node1.Children.Add(node2);
node1.Children.Add(node3);
node1.Children.Add(node4);
//level 2
node2.Children.Add(node5);
node2.Children.Add(node6);
node4.Children.Add(node7);
node4.Children.Add(node8);
node4.Children.Add(node9);
return(tree);
}
public KarthicAryTree BuildAryTreeFromArray(int[] array, int k)
{
//int[] array has value..remember if the value is zero mean there is no node...there might be lot of empty value..
//bcoz all the node in n-ary may or may not have n children
//build the root
List <KarthicAryTreeNode> nodelist = new List <KarthicAryTreeNode>(); //we got to maintain the node list
Dictionary <int, KarthicAryTreeNode> ht = new Dictionary <int, KarthicAryTreeNode>();
//ht key is index and value is tree node
KarthicAryTree tree = new KarthicAryTree();
tree.root = new KarthicAryTreeNode(array[0]);
//nodelist.Add(tree.root);
ht.Add(0, tree.root);
//we already build the root so start from the 1
for (int index = 1; index < array.Length; index++)
{
//only if the value is greater than 0 we build the tree
if (array[index] > 0)
{
//get the parent index
int parentindex = (index - 1) / k;
//since we build level by level the parent node would already be present when the code comes here
//get the parent node
KarthicAryTreeNode parent = ht[parentindex];
//build child
KarthicAryTreeNode child = new KarthicAryTreeNode(array[index]);
//nodelist.Add(child);
ht.Add(index, child);
parent.Children.Add(child);
}
//else
//{
// //Update 5/5/2015
// //If the code comes here mean an array[index] == 0 found
// //that means no children exists for the given index
// //to maintain the index of nodelist so that parent is maintained in right order
// nodelist.Add(null);
//}
}
return(tree);
}
private void button1_Click(object sender, EventArgs e)
{
//Lets buid a n-ary tree
///////////////////1
///////2///////////3//////////////////4
///5//////6//////////////////////7///8////9
///
// For a k-ary tree with height h, the upper bound for the maximum number of leaves is k^h.
//The height h of a k-ary tree does not include the root node, with a tree containing only a root node having a height of 0.
//The total number of nodes in a perfect k-ary tree is (k^{h+1} - 1)/(k-1), while the height h is
KarthicAryTree tree = TreeHelper.SetUpAryTreeWithThreeChildren();
string result = tree.BreathFirstTraversal(tree.root);
//serialize the n-ary tree..convert to array and to bytes
int[] array = ConvertAryTreeIntoString(tree);
string filepath = this.textBox2.Text;
byte[] buffer = GetBytes(array);
//write the output in file
StoreByteSIntofile(filepath, buffer);
//deserialize the tree from byte[] or from file
byte[] bufferfromfile = ConvertFileIntoByte(filepath);
int[] arrayfrombyte = GetArrayFromByte(bufferfromfile);
KarthicAryTree tree2 = BuildAryTreeFromArray(arrayfrombyte, 3);
string result2 = tree2.BreathFirstTraversal(tree2.root);
this.textBox1.Text = result2;
bool output = result.Equals(result2);
}
public int[] ConvertAryTreeIntoString(KarthicAryTree tree)
{
//Logic
//we need to do level by level traversal and store the value in array using below formula
// cth children = k * i + 1 + c
//where k is the maximum number of children the node can have
//i is the index of the parent
//c is the index of the children relative to parent (1st children index o, 2nd children index 1)
// parent index = i - 1 / k
// where i is the index of the child and k is the k-ary number
//how to know the size of the array..
//i guess the interview will give the no of nodes in n-ary tree OR
//the interview won't give you anything
//If the interviewer giving no info about the no of nodes or height, we can calculate no of nodes by height
//No of nodes of any tree (binary or n-ary)
//No. of node = k^(h+1) -1 / (k -1) where k is 2 for binary and h is height of the tree
//calculate the no of nodes using the forumula
// No. of node = k^(h+1) -1 / (k -1)
//This is the same formula for binary tree as well..for binary tree k = 2 so the formula becomes
//2^(h+1) -1
//Get the height of the n-ary tree
int height = tree.GetHeight(tree.root);
//for eg 3-ary tree (0 to 3 children)
// (k^{h+1} - 1) / (k - 1)
int k = 3; //For 3-ary tree
int totalnodes = (int)Math.Ceiling(Math.Pow(k, (height + 1)) - 1) / (k - 1);
int[] array = new int[totalnodes]; //each node can have max of 3 children..so take the max possible size
int index = 0;
array[index] = tree.root.Data;
Queue <AryNodeWithIndex> myqueue = new Queue <AryNodeWithIndex>();
//the root's index is 0
myqueue.Enqueue(new AryNodeWithIndex(tree.root, 0));
while (myqueue.Count != 0)
{
AryNodeWithIndex current = myqueue.Dequeue();
int parentindex = current.index;
//for every parent we start from 0
int childnumber = 0;
foreach (KarthicAryTreeNode child in current.node.Children)
{
int childindex = k * parentindex + 1 + childnumber;
array[childindex] = child.Data;
childnumber += 1;
//visit child
myqueue.Enqueue(new AryNodeWithIndex(child, childindex));
}
}
return(array);
}
