Question

我试图合并两个二叉树，如果我使用“return root”最后，系统告诉我“错误：找不到符号：变量根”。当我添加“return root”时它可以工作对每一个条件。我不明白为什么。此外，在我过去的经验中，似乎我不必在递归函数调用之后编写return语句。有人可以解释一下吗？谢谢！

 public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
    if (t1 == null && t2 == null) {
        return null;
    }
    else if (t1 == null) {
        TreeNode root = new TreeNode(t2.val);
        root.left = mergeTrees(null, t2.left);
        root.right = mergeTrees(null, t2.right);
    }
    else if (t2 == null) {
        TreeNode root = new TreeNode(t1.val);
        root.left = mergeTrees(t1.left, null);
        root.right = mergeTrees(t1.right, null);
    }
    else {
        TreeNode root = new TreeNode(t1.val + t2.val);
        root.left = mergeTrees(t1.left, t2.left);
        root.right = mergeTrees(t1.right, t2.right);
    }
    return root;
}

Answer 1

您的变量root在if，else if和else语句中声明，因此一旦到达if语句块之外的返回，它就会失去作用域。如果你想返回root，我建议你这样做。

     public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
     TreeNode root; //Declare it here so it doesn't lose scope
     if (t1 == null && t2 == null) {
         return null;
     }
     else if (t1 == null) {
         root = new TreeNode(t2.val);
         root.left = mergeTrees(null, t2.left);
         root.right = mergeTrees(null, t2.right);
     }
     else if (t2 == null) {
         root = new TreeNode(t1.val);
         root.left = mergeTrees(t1.left, null);
         root.right = mergeTrees(t1.right, null);
     }
     else {
         root = new TreeNode(t1.val + t2.val);
         root.left = mergeTrees(t1.left, t2.left);
         root.right = mergeTrees(t1.right, t2.right);
     }
         return root;
     }

Answer 2

除非您使用JavaScript编程并使用var，否则大多数algol languages中的变量都具有块范围。因此：

{                                       // new block scope started
  TreeNode root = new TreeNode(t2.val); // New variable TreeNode created
}                                       // Everything created in this scope gets destroyed
root; // Would no compile as root is an unbound variable

如果要创建要返回的变量，则需要在返回的同一范围内创建它。因此：

{
  TreeNode root = null; 
  // .. you logic goes here
  return root; 
} // root goes out of scope here

Trivia：我提到了JavaScript，使用var语言，你的代码可以使用：

function mergeTrees(t1, t2) {
    if (t1 === null && t2 === null) {
        return null;
    }
    else if (t1 === null) {
        var root = new TreeNode(t2.val);
        root.left = mergeTrees(null, t2.left);
        root.right = mergeTrees(null, t2.right);
    } else if (t2 === null) {
        var root = new TreeNode(t1.val);
        root.left = mergeTrees(t1.left, null);
        root.right = mergeTrees(t1.right, null);
    } else {
        var root = new TreeNode(t1.val + t2.val);
        root.left = mergeTrees(t1.left, t2.left);
        root.right = mergeTrees(t1.right, t2.right);
    }
    return root;
}

这样做的原因是var将最近的函数与作用域相关联，而条件中的块不会引入新的作用域。 ES6已使用let和const修复此问题，但其工作方式与其他algol语言相同，并且使用它们的风格更好。

为什么递归函数不需要返回

2 个答案: