本文目錄一覽:
用java怎麼構造一個二叉樹呢?
java構造二叉樹,可以通過鏈表來構造,如下代碼:
public class BinTree {
public final static int MAX=40;
BinTree []elements = new BinTree[MAX];//層次遍歷時保存各個節點
int front;//層次遍歷時隊首
int rear;//層次遍歷時隊尾
private Object data; //數據元數
private BinTree left,right; //指向左,右孩子結點的鏈
public BinTree()
{
}
public BinTree(Object data)
{ //構造有值結點
this.data = data;
left = right = null;
}
public BinTree(Object data,BinTree left,BinTree right)
{ //構造有值結點
this.data = data;
this.left = left;
this.right = right;
}
public String toString()
{
return data.toString();
}
//前序遍歷二叉樹
public static void preOrder(BinTree parent){
if(parent == null)
return;
System.out.print(parent.data+” “);
preOrder(parent.left);
preOrder(parent.right);
}
//中序遍歷二叉樹
public void inOrder(BinTree parent){
if(parent == null)
return;
inOrder(parent.left);
System.out.print(parent.data+” “);
inOrder(parent.right);
}
//後序遍歷二叉樹
public void postOrder(BinTree parent){
if(parent == null)
return;
postOrder(parent.left);
postOrder(parent.right);
System.out.print(parent.data+” “);
}
// 層次遍歷二叉樹
public void LayerOrder(BinTree parent)
{
elements[0]=parent;
front=0;rear=1;
while(frontrear)
{
try
{
if(elements[front].data!=null)
{
System.out.print(elements[front].data + ” “);
if(elements[front].left!=null)
elements[rear++]=elements[front].left;
if(elements[front].right!=null)
elements[rear++]=elements[front].right;
front++;
}
}catch(Exception e){break;}
}
}
//返回樹的葉節點個數
public int leaves()
{
if(this == null)
return 0;
if(left == nullright == null)
return 1;
return (left == null ? 0 : left.leaves())+(right == null ? 0 : right.leaves());
}
//結果返回樹的高度
public int height()
{
int heightOfTree;
if(this == null)
return -1;
int leftHeight = (left == null ? 0 : left.height());
int rightHeight = (right == null ? 0 : right.height());
heightOfTree = leftHeightrightHeight?rightHeight:leftHeight;
return 1 + heightOfTree;
}
//如果對象不在樹中,結果返回-1;否則結果返回該對象在樹中所處的層次,規定根節點為第一層
public int level(Object object)
{
int levelInTree;
if(this == null)
return -1;
if(object == data)
return 1;//規定根節點為第一層
int leftLevel = (left == null?-1:left.level(object));
int rightLevel = (right == null?-1:right.level(object));
if(leftLevel0rightLevel0)
return -1;
levelInTree = leftLevelrightLevel?rightLevel:leftLevel;
return 1+levelInTree;
}
//將樹中的每個節點的孩子對換位置
public void reflect()
{
if(this == null)
return;
if(left != null)
left.reflect();
if(right != null)
right.reflect();
BinTree temp = left;
left = right;
right = temp;
}
// 將樹中的所有節點移走,並輸出移走的節點
public void defoliate()
{
if(this == null)
return;
//若本節點是葉節點,則將其移走
if(left==nullright == null)
{
System.out.print(this + ” “);
data = null;
return;
}
//移走左子樹若其存在
if(left!=null){
left.defoliate();
left = null;
}
//移走本節點,放在中間表示中跟移走…
String innerNode += this + ” “;
data = null;
//移走右子樹若其存在
if(right!=null){
right.defoliate();
right = null;
}
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
BinTree e = new BinTree(“E”);
BinTree g = new BinTree(“G”);
BinTree h = new BinTree(“H”);
BinTree i = new BinTree(“I”);
BinTree d = new BinTree(“D”,null,g);
BinTree f = new BinTree(“F”,h,i);
BinTree b = new BinTree(“B”,d,e);
BinTree c = new BinTree(“C”,f,null);
BinTree tree = new BinTree(“A”,b,c);
System.out.println(“前序遍歷二叉樹結果: “);
tree.preOrder(tree);
System.out.println();
System.out.println(“中序遍歷二叉樹結果: “);
tree.inOrder(tree);
System.out.println();
System.out.println(“後序遍歷二叉樹結果: “);
tree.postOrder(tree);
System.out.println();
System.out.println(“層次遍歷二叉樹結果: “);
tree.LayerOrder(tree);
System.out.println();
System.out.println(“F所在的層次: “+tree.level(“F”));
System.out.println(“這棵二叉樹的高度: “+tree.height());
System.out.println(“————————————–“);
tree.reflect();
System.out.println(“交換每個節點的孩子節點後……”);
System.out.println(“前序遍歷二叉樹結果: “);
tree.preOrder(tree);
System.out.println();
System.out.println(“中序遍歷二叉樹結果: “);
tree.inOrder(tree);
System.out.println();
System.out.println(“後序遍歷二叉樹結果: “);
tree.postOrder(tree);
System.out.println();
System.out.println(“層次遍歷二叉樹結果: “);
tree.LayerOrder(tree);
System.out.println();
System.out.println(“F所在的層次: “+tree.level(“F”));
System.out.println(“這棵二叉樹的高度: “+tree.height());
}
hashmap鏈表大於多少後成為紅黑樹
java8不是用紅黑樹來管理hashmap,而是在hash值相同的情況下(且重複數量大於8),用紅黑樹來管理數據。 紅黑樹相當於排序數據。可以自動的使用二分法進行定位。性能較高。
一般情況下,hash值做的比較好的話基本上用不到紅黑樹。
java中沒有指針,怎麼構建一個鏈表樹
java中要用到鏈表結構的話有LinkedList、LinkedHashSet和LinkedHashMap等集合可供使用,這些集合的底層都是由鏈表實現的
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