本文介绍了如何实现广度优先遍历?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这就是我所拥有的。我认为预订是相同的,并首先将其与深度混合!
This is what I have. I thought pre-order was the same and mixed it up with depth first!
import java.util.LinkedList;
import java.util.Queue;
public class Exercise25_1 {
public static void main(String[] args) {
BinaryTree tree = new BinaryTree(new Integer[] {10, 5, 15, 12, 4, 8 });
System.out.print("\nInorder: ");
tree.inorder();
System.out.print("\nPreorder: ");
tree.preorder();
System.out.print("\nPostorder: ");
tree.postorder();
//call the breadth method to test it
System.out.print("\nBreadthFirst:");
tree.breadth();
}
}
class BinaryTree {
private TreeNode root;
/** Create a default binary tree */
public BinaryTree() {
}
/** Create a binary tree from an array of objects */
public BinaryTree(Object[] objects) {
for (int i = 0; i < objects.length; i++) {
insert(objects[i]);
}
}
/** Search element o in this binary tree */
public boolean search(Object o) {
return search(o, root);
}
public boolean search(Object o, TreeNode root) {
if (root == null) {
return false;
}
if (root.element.equals(o)) {
return true;
}
else {
return search(o, root.left) || search(o, root.right);
}
}
/** Return the number of nodes in this binary tree */
public int size() {
return size(root);
}
public int size(TreeNode root) {
if (root == null) {
return 0;
}
else {
return 1 + size(root.left) + size(root.right);
}
}
/** Return the depth of this binary tree. Depth is the
* number of the nodes in the longest path of the tree */
public int depth() {
return depth(root);
}
public int depth(TreeNode root) {
if (root == null) {
return 0;
}
else {
return 1 + Math.max(depth(root.left), depth(root.right));
}
}
/** Insert element o into the binary tree
* Return true if the element is inserted successfully */
public boolean insert(Object o) {
if (root == null) {
root = new TreeNode(o); // Create a new root
}
else {
// Locate the parent node
TreeNode parent = null;
TreeNode current = root;
while (current != null) {
if (((Comparable)o).compareTo(current.element) < 0) {
parent = current;
current = current.left;
}
else if (((Comparable)o).compareTo(current.element) > 0) {
parent = current;
current = current.right;
}
else {
return false; // Duplicate node not inserted
}
}
// Create the new node and attach it to the parent node
if (((Comparable)o).compareTo(parent.element) < 0) {
parent.left = new TreeNode(o);
}
else {
parent.right = new TreeNode(o);
}
}
return true; // Element inserted
}
public void breadth() {
breadth(root);
}
// Implement this method to produce a breadth first
// search traversal
public void breadth(TreeNode root){
if (root == null)
return;
System.out.print(root.element + " ");
breadth(root.left);
breadth(root.right);
}
/** Inorder traversal */
public void inorder() {
inorder(root);
}
/** Inorder traversal from a subtree */
private void inorder(TreeNode root) {
if (root == null) {
return;
}
inorder(root.left);
System.out.print(root.element + " ");
inorder(root.right);
}
/** Postorder traversal */
public void postorder() {
postorder(root);
}
/** Postorder traversal from a subtree */
private void postorder(TreeNode root) {
if (root == null) {
return;
}
postorder(root.left);
postorder(root.right);
System.out.print(root.element + " ");
}
/** Preorder traversal */
public void preorder() {
preorder(root);
}
/** Preorder traversal from a subtree */
private void preorder(TreeNode root) {
if (root == null) {
return;
}
System.out.print(root.element + " ");
preorder(root.left);
preorder(root.right);
}
/** Inner class tree node */
private class TreeNode {
Object element;
TreeNode left;
TreeNode right;
public TreeNode(Object o) {
element = o;
}
}
}
推荐答案
广度优先搜索
Queue<TreeNode> queue = new LinkedList<BinaryTree.TreeNode>() ;
public void breadth(TreeNode root) {
if (root == null)
return;
queue.clear();
queue.add(root);
while(!queue.isEmpty()){
TreeNode node = queue.remove();
System.out.print(node.element + " ");
if(node.left != null) queue.add(node.left);
if(node.right != null) queue.add(node.right);
}
}
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