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,

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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);

}

}