用java实现二叉树的前序、后序和内序
我正在学习计算机科学的第二学期,在我的数据结构课上,我们看到了递归二叉树。我们必须基于以下Java代码,使用递归进行前序后序和顺序遍历:用java实现二叉树的前序、后序和内序,java,recursion,binary-tree,Java,Recursion,Binary Tree,我正在学习计算机科学的第二学期,在我的数据结构课上,我们看到了递归二叉树。我们必须基于以下Java代码,使用递归进行前序后序和顺序遍历: public class BinaryTree { Node root; public void addNode(int key, String name) { // Create a new Node and initialize it Node newNode = new Node(key, na
public class BinaryTree {
Node root;
public void addNode(int key, String name) {
// Create a new Node and initialize it
Node newNode = new Node(key, name);
// If there is no root this becomes root
if (root == null) {
root = newNode;
} else {
// Set root as the Node we will start
// with as we traverse the tree
Node focusNode = root;
// Future parent for our new Node
Node parent;
while (true) {
// root is the top parent so we start
// there
parent = focusNode;
// Check if the new node should go on
// the left side of the parent node
if (key < focusNode.key) {
// Switch focus to the left child
focusNode = focusNode.leftChild;
// If the left child has no children
if (focusNode == null) {
// then place the new node on the left of it
parent.leftChild = newNode;
return; // All Done
}
} else { // If we get here put the node on the right
focusNode = focusNode.rightChild;
// If the right child has no children
if (focusNode == null) {
// then place the new node on the right of it
parent.rightChild = newNode;
return; // All Done
}
}
}
}
}
// Traversals recursion methods
public void inOrderTraverseTree(Node focusNode) {
}
public void preorderTraverseTree(Node focusNode) {
}
public void postOrderTraverseTree(Node focusNode) {
}
//******************************************************************
public Node findNode(int key) {
// Start at the top of the tree
Node focusNode = root;
// While we haven't found the Node
// keep looking
while (focusNode.key != key) {
// If we should search to the left
if (key < focusNode.key) {
// Shift the focus Node to the left child
focusNode = focusNode.leftChild;
} else {
// Shift the focus Node to the right child
focusNode = focusNode.rightChild;
}
// The node wasn't found
if (focusNode == null)
return null;
}
return focusNode;
}
public static void main(String[] args) {
BinaryTree theTree = new BinaryTree();
theTree.addNode(50, "Boss");
theTree.addNode(25, "Vice President");
theTree.addNode(15, "Office Manager");
theTree.addNode(30, "Secretary");
theTree.addNode(75, "Sales Manager");
theTree.addNode(85, "Salesman 1");
// Different ways to traverse binary trees
theTree.inOrderTraverseTree(theTree.root);
theTree.preorderTraverseTree(theTree.root);
theTree.postOrderTraverseTree(theTree.root);
// Find the node with key 75
System.out.println("\nNode with the key 75");
System.out.println(theTree.findNode(75));
}
}
class Node {
int key;
String name;
Node leftChild;
Node rightChild;
Node(int key, String name) {
this.key = key;
this.name = name;
}
public String toString() {
return name + " has the key " + key;
/*
* return name + " has the key " + key + "\nLeft Child: " + leftChild +
* "\nRight Child: " + rightChild + "\n";
*/
}
}
公共类二进制树{
节根;
public void addNode(int键,字符串名称){
//创建一个新节点并初始化它
Node newNode=新节点(键、名称);
//如果没有根,则成为根
if(root==null){
根=新节点;
}否则{
//将root设置为我们将启动的节点
//当我们穿过这棵树的时候
节点focusNode=root;
//新节点的未来父节点
节点父节点;
while(true){
//root是顶级父级,因此我们开始
//那里
父节点=焦点节点;
//检查新节点是否应继续运行
//父节点的左侧
if(键
有人能给我解释一下这些trasversals是如何工作的以及如何编码的吗?网上有一些二叉树的可视化效果,所以你可以更好地理解它,但这里有一些我使用的图像
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public void inOrderTraverseTree(Node focusNode) {
if (focusNode != null) {
// Traverse the left node
inOrderTraverseTree(focusNode.leftChild);
// Visit the currently focused on node
System.out.println(focusNode);
// Traverse the right node
inOrderTraverseTree(focusNode.rightChild);
}
}
public void postOrderTraverseTree(Node focusNode) {
if (focusNode != null) {
postOrderTraverseTree(focusNode.leftChild);
postOrderTraverseTree(focusNode.rightChild);
System.out.println(focusNode);
}
}
public void preorderTraverseTree(Node focusNode) {
if (focusNode != null) {
System.out.println(focusNode);
preorderTraverseTree(focusNode.leftChild);
preorderTraverseTree(focusNode.rightChild);
}
}