Java中有没有办法在满足某个条件时结束递归方法？

发布于 2024-11-05 05:13:19 字数 543 浏览 6 评论 0原文

有没有办法在满足条件时突然返回到调用下面方法的方法？类似于循环的中断。

public void createTree(TreeNode tree) throws IOException {
    if (i >= preorder.length) {
        // Leave recursive method
    } else if (preorder[i] == '0') {
        tree.value = '0';
        tree.left = tree.right = null;
        i++;                
    } else {
        tree.value = preorder[i];
        i++;
        tree.left = new TreeNode();
        createTree(tree.left);
        tree.right = new TreeNode();
        createTree(tree.right);
    }
}

谢谢。

原文

Is there a way to abruptly return to the method that calls the method below when the condition is met? Something similar to a break for loops.

public void createTree(TreeNode tree) throws IOException {
    if (i >= preorder.length) {
        // Leave recursive method
    } else if (preorder[i] == '0') {
        tree.value = '0';
        tree.left = tree.right = null;
        i++;                
    } else {
        tree.value = preorder[i];
        i++;
        tree.left = new TreeNode();
        createTree(tree.left);
        tree.right = new TreeNode();
        createTree(tree.right);
    }
}

Thanks.

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执笏见 2024-11-12 05:13:19

您可以简单地让该方法返回。然后递归将向后执行，并执行剩余的语句。

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陌若浮生 2024-11-12 05:13:19

您可以从该方法返回状态：

/**
 * @return false if the recursion has been aborted
 */
public boolean createTree(TreeNode tree) throws IOException {
    if (i >= preorder.length) {
        return false;
    } 
    else if (preorder[i] == '0') {
        tree.value = '0';
        tree.left = tree.right = null;
        i++;                
    } 
    else {
        tree.value = preorder[i];
        i++;
        tree.left = new TreeNode();
        if (!createTree(tree.left)) {
            return false;
        };
        tree.right = new TreeNode();
        if (!createTree(tree.right)) {
            return false;
        }
    }
}

You could return a status from the method :

/**
 * @return false if the recursion has been aborted
 */
public boolean createTree(TreeNode tree) throws IOException {
    if (i >= preorder.length) {
        return false;
    } 
    else if (preorder[i] == '0') {
        tree.value = '0';
        tree.left = tree.right = null;
        i++;                
    } 
    else {
        tree.value = preorder[i];
        i++;
        tree.left = new TreeNode();
        if (!createTree(tree.left)) {
            return false;
        };
        tree.right = new TreeNode();
        if (!createTree(tree.right)) {
            return false;
        }
    }
}

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