[英]Converting an iterative Level Order Traversal (of a BST) to a recursive implementation
我有一个工作程序,使用迭代函数的“首选”方法来打印(完整)二进制树的每个级别(请参见下面的代码),但是我想看看如何实现相同的程序,但是使用递归方法。
尽管我同意通常有人为我编写代码不利于良好的学习; 在没有事先看到可行的实现的情况下,学习如何做某事会有更多的麻烦(因为尚未设置任何先例)。
到目前为止,我有什么(迭代程序):
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Collections;
public class BinaryTreeLevelWise<T extends Comparable<T>> {
/*
* This class represents the individual nodes of the binary tree
* Each node has a left, right pointer of type Node
* and Value to hold the value
*/
class Node<T> {
Node left;
Node right;
T value;
public Node(T value) {
this.value = value;
}
@Override
public String toString() {
return "Node value=" + value + "";
}
}
public static void main(String[] args) {
new BinaryTreeLevelWise().run();
}
/*
* This function inserts an element into the binary tree
*/
public <T> void insert(Node node, T value) {
if (((Comparable<T>) value).compareTo((T) node.value) < 0) {
if (node.left != null) {
insert(node.left, value);
} else {
System.out.println(" Inserted " + value + " to left of "
+ node.value);
node.left = new Node(value);
}
} else if (((Comparable<T>) value).compareTo((T) node.value) > 0) {
if (node.right != null) {
insert(node.right, value);
} else {
System.out.println(" Inserted " + value + " to right of "
+ node.value);
node.right = new Node(value);
}
}
}
public void run() {
Node root = new Node(5);
System.out.println("Building tree with root value " + root.value);
insert(root, 1);
insert(root, 8);
insert(root,-2);
insert(root, 6);
insert(root, 3);
insert(root, 9);
insert(root,-3);
insert(root,-1);
insert(root,-4);
System.out.println("*************\nPrinting the tree level by level");
printLevelWise(root);
}
/*
* This functions uses a list of nodes and prints them level by level,
* assuming a complete binary tree.
*/
public void printLevelWise(Node root) {
List<List<Node>> levels = traverseLevels(root);
int i = 0;
for (List<Node> level : levels) {
System.out.print("Level " + i + ": ");
for (Node node : level) {
System.out.print("node " + node.value + " -> ");
}
System.out.println();
i++;
}
}
/*
* This function traverses the tree and puts all the nodes into a list, level by level
*/
private List<List<Node>> traverseLevels(Node root) {
if (root == null) {
return Collections.emptyList();
}
List<List<Node>> levels = new LinkedList<>();
Queue<Node> nodes = new LinkedList<>();
nodes.add(root);
while (!nodes.isEmpty()) {
List<Node> level = new ArrayList<>(nodes.size());
levels.add(level);
for (Node node : new ArrayList<>(nodes)) {
level.add(node);
if (node.left != null) {
nodes.add(node.left);
}
if (node.right != null) {
nodes.add(node.right);
}
nodes.poll();
}
}
return levels;
}
}
此代码输出以下内容(我认为这是正确的输出):
Level 0: node 5 ->
Level 1: node 1 -> node 8 ->
Level 2: node -2 -> node 3 -> node 6 -> node 9 ->
Level 3: node -3 -> node -1 ->
Level 4: node -4 ->
关于如何使该程序使用递归方法而不是迭代方法的任何想法?
您可以尝试下面的代码以递归的方式,但是如果您比较复杂度,那么与以递归方式遍历级别的递归方法相比,Queue方法更好。
public void printLevelOrder(Node<T> root)
{
int h = height(root);//Calculate height of the tree
int i;
for (i=1; i<=h; i++)
{
printGivenLevel(root, i);
System.out.println();
}
}
private void printGivenLevel(Node<T> root, int height) {
if (root == null)
return;
if (height == 1)
System.out.print(root.value);
else if (height > 1)
{
printGivenLevel(root.left, height-1);
printGivenLevel(root.right, height-1);
}
}
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