二叉树代码java 二叉树代码题太难了
java 构建二叉树
首先我想问为什么要用LinkedList 来建立二叉树呢? LinkedList 是线性表,
成都创新互联公司是一家以重庆网站建设、网页设计、品牌设计、软件运维、成都网站营销、小程序App开发等移动开发为一体互联网公司。已累计为成都办公空间设计等众行业中小客户提供优质的互联网建站和软件开发服务。
树是树形的, 似乎不太合适。
其实也可以用数组完成，而且效率更高.
关键是我觉得你这个输入本身就是一个二叉树啊，
String input = "ABCDE F G";
节点编号从0到8. 层次遍历的话:
对于节点i.
leftChild = input.charAt(2*i+1); //做子树
rightChild = input.charAt(2*i+2);//右子树
如果你要将带有节点信息的树存到LinkedList里面, 先建立一个节点类:
class Node{
public char cValue;
public Node leftChild;
public Node rightChild;
public Node(v){
this.cValue = v;
}
}
然后遍历input,建立各个节点对象.
LinkedList tree = new LinkedList();
for(int i=0;i input.length;i++)
LinkedList.add(new Node(input.charAt(i)));
然后为各个节点设置左右子树:
for(int i=0;iinput.length;i++){
((Node)tree.get(i)).leftChild = (Node)tree.get(2*i+1);
((Node)tree.get(i)).rightChild = (Node)tree.get(2*i+2);
}
这样LinkedList 就存储了整个二叉树. 而第0个元素就是树根，思路大体是这样吧。
java构建二叉树算法
//******************************************************************************************************//
//*****本程序包括简单的二叉树类的实现和前序,中序,后序,层次遍历二叉树算法,*******//
//******以及确定二叉树的高度,制定对象在树中的所处层次以及将树中的左右***********//
//******孩子节点对换位置,返回叶子节点个数删除叶子节点,并输出所删除的叶子节点**//
//*******************************CopyRight By phoenix*******************************************//
//************************************Jan 12,2008*************************************************//
//****************************************************************************************************//
public class BinTree {
public final static int MAX=40;
private Object data; //数据元数
private BinTree left,right; //指向左,右孩子结点的链
BinTree []elements = new BinTree[MAX];//层次遍历时保存各个节点
int front;//层次遍历时队首
int rear;//层次遍历时队尾
public BinTree()
{
}
public BinTree(Object data)
{ //构造有值结点
this.data = data;
left = right = null;
}
public BinTree(Object data,BinTree left,BinTree right)
{ //构造有值结点
this.data = data;
this.left = left;
this.right = right;
}
public String toString()
{
return data.toString();
}//前序遍历二叉树
public static void preOrder(BinTree parent){
if(parent == null)
return;
System.out.print(parent.data+" ");
preOrder(parent.left);
preOrder(parent.right);
}//中序遍历二叉树
public void inOrder(BinTree parent){
if(parent == null)
return;
inOrder(parent.left);
System.out.print(parent.data+" ");
inOrder(parent.right);
}//后序遍历二叉树
public void postOrder(BinTree parent){
if(parent == null)
return;
postOrder(parent.left);
postOrder(parent.right);
System.out.print(parent.data+" ");
}// 层次遍历二叉树
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;}
}
}//返回树的叶节点个数
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());
}//结果返回树的高度
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;
}
//如果对象不在树中,结果返回-1;否则结果返回该对象在树中所处的层次,规定根节点为第一层
public int level(Object object)
{
int levelInTree;
if(this == null)
return -1;
if(object == data)
return 1;//规定根节点为第一层
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;
}
//将树中的每个节点的孩子对换位置
public void reflect()
{
if(this == null)
return;
if(left != null)
left.reflect();
if(right != null)
right.reflect();
BinTree temp = left;
left = right;
right = temp;
}// 将树中的所有节点移走,并输出移走的节点
public void defoliate()
{
String innerNode = "";
if(this == null)
return;
//若本节点是叶节点，则将其移走
if(left==nullright == null)
{
System.out.print(this + " ");
data = null;
return;
}
//移走左子树若其存在
if(left!=null){
left.defoliate();
left = null;
}
//移走本节点，放在中间表示中跟移走...
innerNode += this + " ";
data = null;
//移走右子树若其存在
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("前序遍历二叉树结果: ");
tree.preOrder(tree);
System.out.println();
System.out.println("中序遍历二叉树结果: ");
tree.inOrder(tree);
System.out.println();
System.out.println("后序遍历二叉树结果: ");
tree.postOrder(tree);
System.out.println();
System.out.println("层次遍历二叉树结果: ");
tree.LayerOrder(tree);
System.out.println();
System.out.println("F所在的层次: "+tree.level("F"));
System.out.println("这棵二叉树的高度: "+tree.height());
System.out.println("--------------------------------------");
tree.reflect();
System.out.println("交换每个节点的孩子节点后......");
System.out.println("前序遍历二叉树结果: ");
tree.preOrder(tree);
System.out.println();
System.out.println("中序遍历二叉树结果: ");
tree.inOrder(tree);
System.out.println();
System.out.println("后序遍历二叉树结果: ");
tree.postOrder(tree);
System.out.println();
System.out.println("层次遍历二叉树结果: ");
tree.LayerOrder(tree);
System.out.println();
System.out.println("F所在的层次: "+tree.level("F"));
System.out.println("这棵二叉树的高度: "+tree.height());
}
Java数据结构二叉树深度递归调用算法求内部算法过程详解
二叉树
1
2 3
4 5 6 7
这个二叉树的深度是3，树的深度是最大结点所在的层，这里是3.
应该计算所有结点层数，选择最大的那个。
根据上面的二叉树代码，递归过程是：
f(1)=f(2)+1 f(3) +1 ? f(2) + 1 : f(3) +1
f(2) 跟f(3)计算类似上面，要计算左右结点，然后取大者
所以计算顺序是f(4.left) = 0, f(4.right) = 0
f(4) = f(4.right) + 1 = 1
然后计算f(5.left) = 0,f(5.right) = 0
f(5) = f(5.right) + 1 =1
f(2) = f(5) + 1 =2
f(1.left) 计算完毕，计算f(1.right) f(3) 跟计算f(2)的过程一样。
得到f(3) = f(7) +1 = 2
f(1) = f(3) + 1 =3
if(depleftdepright){
return depleft+1;
}else{
return depright+1;
}
只有left大于right的时候采取left +1，相等是取right
java实现二叉树层次遍历
import java.util.ArrayList;
public class TreeNode {
private TreeNode leftNode;
private TreeNode rightNode;
private String nodeName;
public TreeNode getLeftNode() {
return leftNode;
}
public void setLeftNode(TreeNode leftNode) {
this.leftNode = leftNode;
}
public TreeNode getRightNode() {
return rightNode;
}
public void setRightNode(TreeNode rightNode) {
this.rightNode = rightNode;
}
public String getNodeName() {
return nodeName;
}
public void setNodeName(String nodeName) {
this.nodeName = nodeName;
}
public static int level=0;
public static void findNodeByLevel(ArrayListTreeNode nodes){
if(nodes==null||nodes.size()==0){
return ;
}
level++;
ArrayListTreeNode temp = new ArrayList();
for(TreeNode node:nodes){
System.out.println("第"+level+"层:"+node.getNodeName());
if(node.getLeftNode()!=null){
temp.add(node.getLeftNode());
}
if(node.getRightNode()!=null){
temp.add(node.getRightNode());
}
}
nodes.removeAll(nodes);
findNodeByLevel(temp);
}
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
TreeNode root = new TreeNode();
root.setNodeName("root");
TreeNode node1 = new TreeNode();
node1.setNodeName("node1");
TreeNode node3 = new TreeNode();
node3.setNodeName("node3");
TreeNode node7 = new TreeNode();
node7.setNodeName("node7");
TreeNode node8 = new TreeNode();
node8.setNodeName("node8");
TreeNode node4 = new TreeNode();
node4.setNodeName("node4");
TreeNode node2 = new TreeNode();
node2.setNodeName("node2");
TreeNode node5 = new TreeNode();
node5.setNodeName("node5");
TreeNode node6 = new TreeNode();
node6.setNodeName("node6");
root.setLeftNode(node1);
node1.setLeftNode(node3);
node3.setLeftNode(node7);
node3.setRightNode(node8);
node1.setRightNode(node4);
root.setRightNode(node2);
node2.setLeftNode(node5);
node2.setRightNode(node6);
ArrayListTreeNode nodes = new ArrayListTreeNode();
nodes.add(root);
findNodeByLevel(nodes);
}
}
分享标题：二叉树代码java 二叉树代码题太难了
本文路径：http://cdiso.cn/article/dopsgjh.html
