Latest Tutorials| Questions and Answers|Ask Questions?|Site Map



Home Answers Viewqa Java-Beginners How to create binary search tree using an array?

Have Programming Question? Ask it here!
 
 
 


og sim
How to create binary search tree using an array?
1 Answer(s)      2 years and 9 months ago
Posted in : Java Beginners


hello people,

pls guide me on the topic above.

i have an string array, i want to make a binary search tree based on data inside this array.

the array contains names of people, so the tree must have a to l as left child, root as m and right child as n-z, am i right? how do i do it?


Advertisement
View Answers

October 24, 2011 at 12:25 PM


class bnode {   
  String key;
  bnode left;
  bnode right;
  bnode() {           
    key = null;
    left = null;
    right = null;
  }
  bnode(String key) {
    this.key = key;
    left = null;
    right = null;
  }
};
class BinarySearchTree {                
  bnode root;
  BinarySearchTree() {
    root = null;
  }
  void put(String key) {
    bnode current = root;
    bnode prev = current;
    if (root == null) {
      root = new bnode(key);
    }
    else {
      boolean insert = false;
      while (insert == false) {
        prev = current;
        if (key.compareTo(current.key) < 0) {
          if (current.left == null) {
            current.left = new bnode(key);
            insert = true;
          }
          current = current.left;
        }
        else {
          if (current.right == null) {
            current.right = new bnode(key);
            insert = true;
          }
          current = current.right;
        }
      }
    }
  }
  boolean delete(String key) {        
    boolean deleted = true;
    bnode current = root;
    bnode prev = current;
    while (current != null) {
      if (key.compareTo(current.key) > 0) {
        prev = current;
        current = current.right;
      }
      else if (key.compareTo(current.key) < 0) {
        prev = current;
        current = current.left;
      }
      else if (key.compareTo(current.key) == 0) {
        deleted = false;
        break;
      }
    }
    if (check(current) == 0) {
      if (current.key.compareTo(prev.key) > 0) {
        prev.right = null;
      }
      else {
        prev.left = null;
      }
    }
    else if (check(current) == 1) {

      if (current.key.compareTo(prev.key) > 0) {
        if (current.left != null) {
          prev.right = current.left;
        }
        else {
          prev.right = current.right;
        }
      }
      else {
        if (current.left != null) {
          prev.left = current.left;
        }
        else {
          prev.left = current.right;
        }
      }
    }
    else if (check(current) == 2) {
      bnode temp = inord(current);
      if (current == root) {
        root.key = temp.key;
      }
      else {
        if (current.key.compareTo(prev.key) > 0) {
          prev.right.key = temp.key;
        }
        else {
          prev.left.key = temp.key;
        }
      }
    }
    return deleted;
  }
  bnode inord(bnode a) {                    
    int t = 0;
    bnode ret, prev = new bnode();
    prev = a;
    a = a.right;
    while (a.left != null) {
      prev = a;
      a = a.left;
      t = 1;
    }
    ret = a;
    if (t == 0) {
      prev.right = null;
    }
    else {
      prev.left = null;
    }
    a = null;
    return ret;
  }
  int check(bnode a) { 
    int ret;
    if ( (a.left != null) && (a.right != null)) {
      ret = 2;
    }
    else if ( (a.left == null) && (a.right == null)) {
      ret = 0;
    }
    else {
      ret = 1;
    }
    return ret;
  }
  void printIn(bnode oot) {                  
    if (oot.left != null) {
      printIn(oot.left);
    }
    System.out.println("--------" + oot.key + "----------");
    if (oot.right != null) {
      printIn(oot.right);
    }
  }
  public static void main(String[] args) {
    BinarySearchTree a = new BinarySearchTree();
    String names[]={"A","B","C","D"};
    for(int i=0;i<names.length;i++){
        a.put(names[i]);
    }
    a.printIn(a.root);
  }
}


Related Tutorials/Questions & Answers:
Advertisements
Advertisements
 

 

 

DMCA.com