This program demonstrates a few routines for processing binary sort trees. It uses a binary sort tree of strings. The user types in strings. The user’s string is converted to lower case, and — if it is not already in the tree — it is inserted into the tree. Then the number of nodes in the tree and a list of items in the tree are output. The program ends when the user enters an empty string.

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public class SortTreeDemo {

   static class TreeNode {
           // An object of type TreeNode represents one node
           // in a binary tree of strings.
      String item;      // The data in this node.
      TreeNode left;    // Pointer to left subtree.
      TreeNode right;   // Pointer to right subtree.
      TreeNode(String str) {
             // Constructor.  Make a node containing the specified string.
         item = str;
   }  // end nested class TreeNode

   static TreeNode root;  // Pointer to the root node in a binary tree.  This
                          // tree is used in this program as a binary sort tree.
                          // The tree is not allowed to contain duplicate
                          // items.  When the tree is empty, root is null.

   public static void main(String[] args) {

      TextIO.putln("This programs stores strings that you enter in a binary sort");
      TextIO.putln("tree.  After each items is inserted, the contents of the tree");
      TextIO.putln("are displayed.  The number of nodes in the tree is also output.");
      TextIO.putln("    Any string you enter will be converted to lower case.");
      TextIO.putln("Duplicate entries are ignored.");

      while (true) {
              // Get one string from the user, insert it into the tree,
              // and print some information about the tree.  Exit if the
              // user enters an empty string.  Note that all strings are
              // converted to lower case.
          TextIO.putln("\n\nEnter a string to be inserted, or press return to end.");
          TextIO.put("?  ");
          String item;  // The user's input.
          item = TextIO.getln().trim().toLowerCase();
          if (item.length() == 0)
          if (treeContains(root,item)) {
                 // Don't insert a second copy of an item that is already
                 // in the tree.
             TextIO.putln("\nThat item is already in the tree.");
          else {
             treeInsert(item);  // Add user's string to the tree.
             TextIO.putln("\nThe tree contains " + countNodes(root) + " items.");
             TextIO.putln("\nContents of tree:\n");
      }  // end while

      TextIO.putln("\n\nExiting program.");

   }  // end main()

   static void treeInsert(String newItem) {
          // Add the item to the binary sort tree to which the global
          // variable "root" refers.  (Note that root can't be passed as
          // a parameter to this routine because the value of root might
          // change, and a change in the value of a formal parameter does
          // not change the actual parameter.)
      if ( root == null ) {
             // The tree is empty.  Set root to point to a new node containing
             // the new item.  This becomes the only node in the tree.
         root = new TreeNode( newItem );
      TreeNode runner;  // Runs down the tree to find a place for newItem.
      runner = root;   // Start at the root.
      while (true) {
         if ( newItem.compareTo(runner.item) < 0 ) {
                  // Since the new item is less than the item in runner,
                  // it belongs in the left subtree of runner.  If there
                  // is an open space at runner.left, add a new node there.
                  // Otherwise, advance runner down one level to the left.
            if ( runner.left == null ) {
               runner.left = new TreeNode( newItem );
               return;  // New item has been added to the tree.
               runner = runner.left;
         else {
                  // Since the new item is greater than or equal to the item in
                  // runner it belongs in the right subtree of runner.  If there
                  // is an open space at runner.right, add a new node there.
                  // Otherwise, advance runner down one level to the right.
            if ( runner.right == null ) {
               runner.right = new TreeNode( newItem );
               return;  // New item has been added to the tree.
               runner = runner.right;
      } // end while
   }  // end treeInsert()

   static boolean treeContains( TreeNode root, String item ) {
          // Return true if item is one of the items in the binary
          // sort tree to which root points.   Return false if not.
      if ( root == null ) {
            // Tree is empty, so it certainly doesn't contain item.
         return false;
      else if ( item.equals(root.item) ) {
            // Yes, the item has been found in the root node.
         return true;
      else if ( item.compareTo(root.item) < 0 ) {
            // If the item occurs, it must be in the left subtree.
         return treeContains( root.left, item );
      else {
            // If the item occurs, it must be in the right subtree.
         return treeContains( root.right, item );
   }  // end treeContains()

   static void treeList(TreeNode node) {
          // Print the items in the tree in postorder, one item
          // to a line.  Since the tree is a sort tree, the output
          // will be in increasing order.
      if ( node != null ) {
          treeList(node.left);             // Print items in left subtree.
          TextIO.putln("  " + node.item);  // Print item in the node.
          treeList(node.right);            // Print items in the right subtree.
   } // end treeList()

   static int countNodes(TreeNode node) {
         // Count the nodes in the binary tree to which node
         // points.  Return the answer.
      if ( node == null ) {
              // Tree is empty, so it contains no nodes.
         return 0;
      else {
             // Add up the root node and the nodes in its two subtrees.
         int leftCount = countNodes( node.left );
         int rightCount = countNodes( node.right );
         return  1 + leftCount + rightCount;
   } // end countNodes()

} // end class SortTreeDemo