打印二叉树的所有路径

Question

我正在研究打印二叉树的所有路径的问题，这会产生结果。我创建了一个全局变量 sw ，并且在 printAllRootToLeafPaths 方法的递归中创建了一个String变量正在使用路径 。我是否可以通过某种方式在 printAllRootToLeafPaths > printAllRootToLeafPaths < printAllRootToLeafPaths < em> 方法？因此，该方法将如下所示。

public static ArrayList<String> printAllRootToLeafPaths(TreeNode node ){

   /*
      String path and ArrayList<String> sw will be initiated here 
   */
}

==============================================================================

import java.io.*;
import  java.util.*; 

class TreeNode {

      int val;
      TreeNode left;
      TreeNode right;

      TreeNode(int x) { 
        val = x; 
    }
}


public class myTest {



    public static ArrayList<String> sw = new ArrayList<String>(); 

    public static void main ( String[] args ){

        TreeNode root = new TreeNode( 1 );

        root.left= new TreeNode( 2 ) ;
        root.left.left =  new TreeNode( 5 );

        root.right =  new TreeNode(3);


        sw =  printAllRootToLeafPaths( root, new String() ); 
        String[] result =  new String[ sw.size() ]; 
        int count =  0 ; 

        for ( String s: sw ){

            result[count] = '"'+ s + '"'; 
            count++; 
        }

        System.out.println( Arrays.toString( result ) );

    }


    public static ArrayList<String> printAllRootToLeafPaths(TreeNode node, String path ) {

        if( node==null ) return null ; 

        path += String.valueOf(node.val)+ "->"; 

        if( node.left == null && node.right == null ){

            String my = path.substring(0, path.length() -2 ); 
            sw.add( my );

            // optional 
            path = ""; 
        }

        else {

            printAllRootToLeafPaths( node.left, new String (path) );
            printAllRootToLeafPaths( node.right, new String (path)  );
        }  

        return sw ;     
    }

}

Answer 1

sw =   printAllRootToLeafPaths(node.left, new String());
sw =   printAllRootToLeafPaths(node.right, new String());

当你传递路径时（而不是new String()是你在所有方法调用中使用单个对象，这意味着，当你返回到原始调用者时，该对象与它的状态不同然后。调用递归方法

Answer 2

看看这个解决方案：

switch

如您所见，您只需要printAllRootToLeafPaths中的路径字符串。但是，这个函数有两个参数。您可以替换全局变量的第二个参数，但是使用递归会更难维护。

此代码的结果是：

class TreeNode {

    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
        val = x;
    }
}
public class myTest {

    public static void main(String[] args) {

        TreeNode root = new TreeNode(1);

        root.left = new TreeNode(2);
        root.left.left = new TreeNode(5);

        root.right = new TreeNode(3);

        printAllRootToLeafPaths(root, new String());

    }

    public static void printAllRootToLeafPaths(TreeNode node, String path) {
        path = path + " -> " + node.val;
        if (isLeaf(node)) {

            System.out.println(path);
        } else if (node.left == null && node.right != null) {
            printAllRootToLeafPaths(node.right, path);
        } else if (node.left != null && node.right == null) {
            printAllRootToLeafPaths(node.left, path);
        } else {
            printAllRootToLeafPaths(node.left, path);
            printAllRootToLeafPaths(node.right, path);
        }

    }

    public static boolean isLeaf(TreeNode t) {
        if (t.left == null && t.right == null) {
            return true;
        }
        return false;
    }

}

Answer 3

  public static List<String> printAllRootToLeafPaths(TreeNode node ){

       /*
          String path and ArrayList<String> sw will be initiated here 
       */
     List<String> sw =new ArrayList<String>();
     StringBuilder path=new StringBuilder();
     printAllRootToLeafPaths(TreeNode node,path,sw ) ;
    }

如果函数不是静态的，则函数需要将引用传递给每个调用中的列表

   public static List<String> printAllRootToLeafPaths(TreeNode node, StringBuilder path,List<String> pathList ) 
    {
     ...
    else {
           printAllRootToLeafPaths( node.left, new StringBuilder (path.toString()),pathList );
           printAllRootToLeafPaths( node.right, new StringBuilder (path.toString()),pathList  );
            }