如何通过java中的递归返回值列表? [英] How to return a list of values through recursion in java?
问题描述
通过递归返回单个值效果很好.但是如果我想返回一个递归遍历每个调用的值列表怎么办.这是我的代码.
Returning a single value through recursion works just fine. But what if I want to return a list of values that recursion goes through each call. Here's my code.
public void inOrder(Node focusNode) {
/* ArrayList<Integer> tempList = new ArrayList<Integer>(); */
if ( focusNode != null) {
inOrder(focusNode.getLeftNode());
System.out.println(focusNode);
/* tempList.add(focusNode.getElement()); */
inOrder(focusNode.getRightNode());
}
/* int[] elems = new int[tempList.toArray().length];
int i = 0;
for ( Object o : tempList.toArray())
elems[i++] = Integer.parseInt(o.toString()); */
//return tempList;
}
遍历时打印值给出了预期的输出.但是存储这些值不起作用.它只返回一个列表中的单个值.有人可以帮我解决这个问题吗?
printing values while traversing through gives the expected output. But storing those values is not working. It only returns with a single value in a list. Can someone help me with this?
推荐答案
为什么不直接传入对数组列表的引用以及起始节点?在您的 inOrder 方法运行后,您将拥有一系列有序的值,您可以随意使用.
Why don't you just pass in a reference to an array list along with your starting node? After your inOrder method runs, you'll have an ordered series of values that you can use as you please.
// method signature changed
public void inOrder(Node focusNode, ArrayList vals) {
/* ArrayList<Integer> tempList = new ArrayList<Integer>(); */
if ( focusNode != null) {
// args changed here
inOrder(focusNode.getLeftNode(), vals);
// adding node to array list rather than dumping to console
vals.add(focusNode);
/* tempList.add(focusNode.getElement()); */
inOrder(focusNode.getRightNode());
}
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