高手指点二元高度(binary height)

大家好:
我在看<<oracle database 10g performance tuning tips & techniques>>时,看到索引那一章时,关于索引的二度高度,讲的不够细致,请高手指点一下,为什么删除了表中的20%-25%的数据后会增加二度高度的值?
还有直方图的那一节也讲的不够详细,谁能详细的给讲一下直方图创建后是如何提高的性能,如果能细致的讲一下,更是感激不尽.
谢谢各位大侠了!

为什么没有回答呢?是不是我问的不够详细,大家没有看懂呀?

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Having trouble finding the height of a Binary Tree

Hi, I have an ADT class called DigitalTree that uses Nodes to form a binary tree; each subtree only has two children at most. Each node has a "key" that is just a long value and is placed in the correct position on the tree determined by its binary values. For the height, I'm having trouble getting an accurate height. With the data I'm using, I should get a height of 5 (I use an array of 9 values/nodes, in a form that creates a longest path of 5. The data I use is int[] ar = {75, 37, 13, 70, 75, 90, 15, 13, 2, 58, 24} ). Here is my code for the whole tree. If someone could provide some tips or clues to help me obtain the right height value, or if you see anything wrong with my code, it would be greatly aprpeciated. Thanks!
public class DigitalTree<E> implements Copyable
   private Node root;
   private int size;
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       size = 0;
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       try
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       else
           if(parent.right == null)
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               size++;
               //System.out.println(size + " " + k);
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           leftHeight = getHeight(currentNode.left, leftHeight, rightHeight);
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You were on the right track with the part you commented out; first define a few things:
1) the height of an empty tree is nul (0);
2) the height of a tree is one more than the maximum of the heights of the left and right sub-trees.
This translates to Java as a recursive function like this:
int getHeight(Node node) {
   if (node == null) // definition #1
      return 0;
   else // definition #2
      return 1+Math.max(getHeight(node.left), getHeight(node.right));
}kind regards,
Jos

Binary tree - a spesific node's height !

hello !
im looking for a method to deterine a certain node's hight in a binary tree (that is its distance from the root )
please enlighten me on how to do this if you can
thank you

> Traverse the BT and with every step increase a
counter. When you get to the node you're looking for,
return the counter.
Or just add an attribute to you TreeNode class which holds the value for the degree/height of the node.

Getting the height of a binary tree

So I am trying to create a method in my binary tree class that returns and integer that represents the height of the tree.
I tried to do this recursively and it seems the numbers are close to the actual ones, but never exact and I haven't really been able to find out where the problem lies. This is my code:
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j2ee_ubuntu wrote:
it may help at some extent..
private int getMax(int x, int y)
          int result = 0;
          if(x>y)
               result = x;
          else if(y>x)
               result = y;
else //when both are equal
result =x; OR result = y;//You can use one of them
          return result;
     }Edited by: j2ee_ubuntu on Nov 6, 2008 12:30 AMWhy not just use [ Math.max|http://java.sun.com/javase/6/docs/api/java/lang/Math.html#max(int,%20int)] or write
public static int max(int a, int b) {
    return (a >= b) ? a : b;
}

How can i find out the height lenght width of that logo in so10 ?

I have TextName is there in so10.
How can i find out the height lenght width of that logo ?

Hi Sam,
In SE80, you should have a Repository for MIMES, including images.You should find your image here .
you have to use the following method to get the image as a binary string and then use gui_download to download it to local system
call method cl_ssf_xsf_utilities=>get_bds_graphic_as_bmp
exporting
p_object = fs_stxbitmaps-tdobject
p_name = fs_stxbitmaps-tdname
p_id = fs_stxbitmaps-tdid
p_btype = fs_stxbitmaps-tdbtype
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exceptions
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Regards
Naresh

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Hi,
I was just wondering, if there's a binary search tree that holds objects can the binary search tree be based on any of the attributes in those objects.
So for example say we have a Person object and it has the attributes name, age, and height. Can we make a binary search tree based on the age or a search tree based on just the height.
And lastly how would that be done in the comparable method.
Thanks.

Of course it depends on what (and whether) the Person class implements for Comparable. If you want to base the tree on the default comparison (i.e., the one that compareTo does) then you don't need a separate Comparator.
When you write your binary tree implementation, it's wise to make it smart enough to either take a Comparator, or to use the default comparison (if the objects being tracked implement Comparable).
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Binary tree using inorder traversal algorithm

I need to draw a Binary tree using Inorder traversal algorithm.. in java applets using rectangles and ellipse objects.. can anyone help me about this ? I will be really thankful to yu

I think this is my psychic Sunday, so here's my attempt at answering
this vaguely stated question. First have a look at this fine piece of
ASCII art:--------------------------------------------   ^         ^
| A                    T                     |   |         |
|                                            |   |        root
---------L-----------------------R---------+   |         v
| B                    | C                  |   |
|                      |                     |   |
|                      |                     | height
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
|                      |                     |   |
-------------------------------------------+   v
<--------------------width------------------->Suppose you have a rectangle 'width' wide and 'height' high. The root
of the tree can be drawn at position T and (if present), two lines can be
drawn to the positions L and R respectively. Now recursively call your
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right subtrees L and R respectively.
kind regards,
Jos

How to Pretty Print a Binary Tree?

I'm trying to display a Binary Tree in such a way:
________26
___13_________2
1_______4 3_________1
(without the underscores)
however I cannot figure out the display method.
class BinaryNode
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Object data;
     BinaryNode()
leftChild = null;
data = null;
rightChild = null;
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leftChild = left;
data = d;
rightChild = right;
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if (root == null)
return 0;
if ((Height(root.leftChild)) > Height(root.rightChild))
return 1 + Height(root.leftChild);
return 1 + Height(root.rightChild);
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public static int Count(BinaryNode root)
if(root==null)
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return 1 + Count(root.leftChild) + Count(root.rightChild);
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Edited by: 815035 on Nov 23, 2010 12:27 PM

The example of what the OP wants it to look like I thought was quite plain. Its right at the top of the post.
Unfortunately it is also quite difficult to accomplish using System.out.print statements.
You have to print out the root of the tree first (its at the top)
However you don't know how far along to the right you need to print it without traversing the child nodes already (you need to know how deep the tree is, and how far to the left the tree extends from the root)
So you will need to traverse the tree at least twice.
Once to work out the offsets, and again to print out the values.
The working out of offsets would have to be a depth search traversal I think
The printing of the values in this fashion would be a breadth first traversal.
I remember (ages ago) doing a similar assignment, except we printed the tree sideways.
ie the root was on the left, the leaves of the tree on the right of the screen.
That meant you could do an inorder depth traversal of the tree to just print it once.
hope this helps,
evnafets

Recover a Binary Tree

Ok people I need help!
I want to create, from a pre-order and in-order sequences, the post-order sequence! Obviously there are no repeated elements (characters from A to Z at this case).
The tree root it's the first character from pre-order and after that it's easy to take left and right sequences.
I found an interesting explanation: http://www.dcs.kcl.ac.uk/staff/pinz...st/html/l7.html
I've tried everything to make it recursive I think that's the best way to do it but somehow I can't make it work! My last code:
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import java.util.*;
public class Problem
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                        System.out.println();
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                        else if (newInOrdLeft.length() != 0) { tempTree.left = work(preOrd, preOrdPoint+1, newInOrdLeft); }
                catch (NoSuchElementException e) {}
                try
                        String newInOrdRight = tok.nextToken();
                        if (newInOrdRight.length() == 1) { tempTree.right = new BinaryNode(newInOrdRight.toCharArray()[0]); }
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class BinaryNode
        char chr;
        BinaryNode left = null;
        BinaryNode right = null;
        BinaryNode()
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                left = lt;
                right = rt;
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}

I don't know if this is a translation of your C code, but it can reconstruct any binary tree (with letters as nodes and no duplicates) from preorder and inorder traversals. Hope it helps.
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public class ReconstructTree {
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        Map table2 = new HashMap();
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        printPostOrder(root, table2);
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digits[i] = string[
temp / binaryPlaceValues[i]];
temp = temp - binaryPlaceValues[i];
} else {
digits[i] = string[0];
public override function create(): Node {
for( i in [0..16]) {
insert string[0] into digits;
for (i in [0..16]) {
insert Text {
font: Font {
size: 24
x: 10,
y: 30
content: bind digits[i]
} into digitNodes;
return Group {
content: HBox{
spacing: 10
content:[
digitNodes

As promised, completed with all four major bases.
import javafx.animation.Interpolator;
import javafx.animation.Timeline;
import javafx.scene.CustomNode;
import javafx.scene.Group;
import javafx.scene.layout.HBox;
import javafx.scene.layout.VBox;
import javafx.scene.Node;
import javafx.scene.Scene;
import javafx.scene.text.Font;
import javafx.scene.text.Text;
import javafx.stage.Stage;
* @author Sidney Harrell
def binaryCounter = BinaryCounter{};
def octalCounter = OctalCounter{};
def decimalCounter = DecimalCounter{};
def hexCounter = HexCounter{};
var time: Integer on replace OldValue{
    binaryCounter.flip(time);
    octalCounter.flip(time);
    decimalCounter.flip(time);
    hexCounter.flip(time);
def timer = Timeline {
    keyFrames: [
        at (0s) {time => 0},
        at (6000s) {time => 100000 tween Interpolator.LINEAR }
Stage {
    title: "Counters"
    width: 530
    height: 200
    scene: Scene {
        content: VBox {
            spacing: 10;
            content: [
                binaryCounter
                octalCounter
                decimalCounter
                hexCounter
timer.play();
abstract class Counter extends CustomNode {
    def digits = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9",
        "A", "B", "C", "D", "E", "F"];
    var placeValues: Integer[];
    var representation: String[];
    var digitNodes: Text[];
    var temp: Integer;
    var base: Integer;
    var places: Integer;
    public function flip(time:Integer) {
        if (time > placeValues[0] * base) {
            temp=time mod placeValues[0] * base;
        } else {
            temp=time;
        for (i in [0..places]) {
            if ((temp != 0)and(temp / placeValues[i] != 0)) {
                representation[i] = digits[temp / placeValues];
temp = temp - placeValues[i]*(temp/placeValues[i]);
} else {
representation[i] = digits[0];
class OctalCounter extends Counter {
public override function create(): Node {
base = 8;
places = 9;
placeValues = [134217728, 16777216, 2097152, 262144, 32768,
4096, 512, 64, 8, 1];
for( i in [0..places]) {insert digits[0] into representation;};
for (i in [0..6]) {insert Text {font: Font {size: 24}; x: 10, y: 30 content: digits[0]} into digitNodes;};
for (i in [0..places]) {insert Text {font: Font {size: 24}; x: 10, y: 30 content: bind representation[i]} into digitNodes;};
insert Text {font: Font {size: 24}; x: 10, y: 30 content: "octal"} into digitNodes;
return Group {content: HBox{spacing: 10 content:[digitNodes];}};
class BinaryCounter extends Counter {
public override function create(): Node {
base = 2;
places = 16;
placeValues = [16384 * 2 * 2, 16384 * 2, 16384, 8192, 4096, 2048, 1024,
512, 256, 128, 64, 32, 16, 8, 4, 2, 1];
for( i in [0..places]) {insert digits[0] into representation;};
for (i in [0..places]) {insert Text {font: Font {size: 24}; x: 10, y: 30 content: bind representation[i]} into digitNodes;};
insert Text {font: Font {size: 24}; x: 10, y: 30 content: "binary"} into digitNodes;
return Group {content: HBox{spacing: 10 content:[digitNodes];}};
class DecimalCounter extends Counter {
public override function create(): Node {
base = 10;
places = 8;
placeValues = [100000000, 10000000, 1000000, 100000, 10000, 1000,
100, 10, 1];
for( i in [0..places]) {insert digits[0] into representation;};
for (i in [0..7]) {insert Text {font: Font {size: 24}; x: 10, y: 30 content: digits[0]} into digitNodes;};
for (i in [0..places]) {insert Text {font: Font {size: 24}; x: 10, y: 30 content: bind representation[i]} into digitNodes;};
insert Text {font: Font {size: 24}; x: 10, y: 30 content: "decimal"} into digitNodes;
return Group {content: HBox{spacing: 10 content:[digitNodes];}};
class HexCounter extends Counter {
public override function create(): Node {
base = 16;
places = 6;
placeValues = [16777216, 1048576, 65536, 4096, 256, 16, 1];
for( i in [0..places]) {insert digits[0] into representation;};
for (i in [0..9]) {insert Text {font: Font {size: 24}; x: 10, y: 30 content: digits[0]} into digitNodes;};
for (i in [0..places]) {insert Text {font: Font {size: 24}; x: 10, y: 30 content: bind representation[i]} into digitNodes;};
insert Text {font: Font {size: 24}; x: 10, y: 30 content: "hex"} into digitNodes;
return Group {content: HBox{spacing: 10 content:[digitNodes];}}
Edited by: sidney.harrell on Jan 12, 2009 7:49 PM
added missing closing brace

Writting Methods of Binary Tree

I am using java code to write some binary tree methods.
Got stuck in some of the following methods.
Hopefully i can get some help here.
Public int width(){
// Return the maximun number of nodes in each level
Public boolean checkIsFullTree(){
// Return true if the tree is a full tree (If all leaves are at the same depth then the tree is full)
Public boolean checkIsCompleteTree(){
// Return true if the tree is a full tree (If all leaves are at the same depth then the tree is full and is filled from left to right)
at last
Public int internalPathLength(){
// Return The sum of all internal nodes of the paths from the root of an extended binary tree to each node.
other words, the sum of all depths of nodes.
Looking forawad to see some replies. Chrees

I have came up with some ideas.
Public boolean checkIsFullTree(TreeNode t, int currentHeight ) {
int maxHeight = height(); // Maximum height.
int currentHeight=currentHeight;
if (t == null){
return false;
if (t.left() ! = null){
currentHeight++;
checkIsFullTree(t.left(), currentHeight);
if (t.right() ! = null){
currentHeight++;
checkIsFullTree(t.right(), currentHeight);
if ( t.left() == null && t.right() == null){
if (currentHeight == maxHeight) {
check = true;
}else{
check = false;
But i am missing sometime here i guess.
}

高手指点二元高度(binary height)

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