binary-tree

I was asked to write the iterative version, but I wrote the recursive version i.e. void inorderTraverse(BinaryTree root) { if(root==NULL) printf("%d",root->id); else { inorderTraverse(root->left); printf("%d",root->id); inorderTraverse(root->right); } } I'm not looking for the code, I want to understand how this can be done. Had it been just the last recursive call, I would have done void inorderTraverse(BinaryTree root) { while(root!=NULL) { printf("%d",root->id); root=root->right; } } But how do I convert to an iterative program when there are two recursive calls? Here are the type

i wanted to know how to read values from a list into a binary tree. i have a triangle like this: 0 1 2 3 4 5 6 7 8 9 i have written a class node like this class node: def __init__(self,data,left=None,right=None): self.data=data self.left=left self.right=right basically what i want to do is something like this node(0,node(1),node(2)) i want to make a recursive function that can handle much bigger triangles. Can somehow tell me what i am supposed to do? edit: quite clearly binary tree is not the way to approach this problem. what i basically want to find out are all the different combination's

I would like to free memory from my allocated binary tree what traversal is the best for doing so? typedef struct Node{ struct Node * right; struct Node * left; void * data; }Node; typedef int (*cmp) (void*,void *); Node* init(void * element){ Node * newNode=(Node*)malloc(sizeof(Node)); newNode->data=element; newNode->left=NULL; newNode->right=NULL; return newNode; } void insert(void * element, Node** root,cmp compareTo){ if(*root==NULL){ *root=init(element); return; } if(compareTo(element,(*root)->data)==1) insert(element,&((*root)->left),compareTo); else insert(element,&((*root)->right)

How can I find the preorder listing of a tree if only the postorder listing is given and vice versa. Also, in the tree, every non-leaf node has two children (i.e. Each node has either two or zero children.) EDIT: Another given assumption is that the label of each node is unique and has a field that will identify it as an internal node or a leaf. I think that ought to get rid of the ambiguity of a single preorder or postorder being able to uniquely identify a tree. Without assuming that nodes in the tree have a field that idenrify themselves as internal or leaf, you can't find a unique answer

I have a programming assignment that goes like this: You are given three numbers a, b, and c. (1 ≤ а, b, c ≤ 10^18) Each time you have two choises, either add b to a (a+=b), or add a to b (b+=a). Write a program that will print out YES or NO depending on whether you can get to c by adding a and b to each other. I've tried solving this problem using recursion that branches to two branches every time where one branch stores a+b, b and the other branch stores a, b+a. In every recursive call, the function checks the values of a and b, and if they are equal to c the search stops and the function

Background I have many (thousands!) of data files with a standard field based format (think tab-delimited, same fields in every line, in every file). I'm debating various ways of making this data available / searchable. (Some options include RDBMS, NoSQL stuff, using the grep/awk and friends, etc.). Proposal In particular, one idea that appeals to me is "indexing" the files in some way. Since these files are read-only (and static), I was imagining some persistent files containing binary trees (one for each indexed field, just like in other data stores). I'm open to ideas about how to this, or

path = 0 # the lenght of the path while self.right != None or self.left != None: while self.right != None: self = self.right path = path +1 while self.left != None: self = self.left path = path +1 return path this is my sample code for find the Height, is defined as the length of the longest path by number of nodes from self to a leaf. The height of a leaf node is 1. it doesn't work. What you're doing isn't recursive, it's iterative. Recursive would be something like: def height(node): if node is None: return 0 else: return max(height(node.left), height(node.right)) + 1 You were given the

Using the method presented here: http://cslibrary.stanford.edu/110/BinaryTrees.html#java 12. countTrees() Solution (Java) /** For the key values 1...numKeys, how many structurally unique binary search trees are possible that store those keys? Strategy: consider that each value could be the root. Recursively find the size of the left and right subtrees. */ public static int countTrees(int numKeys) { if (numKeys <=1) { return(1); } else { // there will be one value at the root, with whatever remains // on the left and right each forming their own subtrees. // Iterate through all the values that

I want to draw it like this Image : I can draw a binary tree on console. I want to draw it using WPF. Here is my code which I write for console. class Program { static void Main(string[] args) { List<BinaryTreeData> myBinaryData = new List<BinaryTreeData>(); myBinaryData.Add(new BinaryTreeData{ownID=1}); myBinaryData.Add(new BinaryTreeData { parentID=1, ownID = 2 }); myBinaryData.Add(new BinaryTreeData { parentID=1,ownID = 3 }); foreach (var item in myBinaryData) { Console.WriteLine("{0}------{1}", item.parentID, item.ownID); } } } class BinaryTreeData : INotifyPropertyChanged { private int