java - 二叉树的遍历问题

Question

最近在做leetcode上的题目，有一道题是要求交换二叉树左右子树。我一开始没有在函数体中加入如下代码：

          if(root == null)
             return null;

结果发现出现空指针异常。我觉得上面这段代码有点多余，但是OJ缺了它通过不了。麻烦各位大神帮忙解决一下小弟的困惑。下面贴上整个程序的代码：

public class TreeNode {
          int val;
          TreeNode left;
          TreeNode right;
          TreeNode(int x) { val = x; }
     
     public TreeNode invertTree(TreeNode root){
         if(root == null)
             return null;
         if(root.left == null && root.right == null)
             return root;
         
         TreeNode temp = root.left;
         root.left = root.right;
         root.right = temp;
         
         if(root.left != null)
             invertTree(root.left);
         
         if(root.right != null)
             invertTree(root.right);
         
         return root;
     }
}

Answer 1

Will his use case directly pass in null? invertTree(null);

Answer 2

You add code, the first code is
if(root.left == null && root.right == null)

         return root;

If the passed parameter is null, then root.left. Because root is null, calling the left attribute will report an exception
If you block the exception, then .root.left==null is true.
I don’t I haven’t understood his code, but I guess the code should be like this.

try{
invertTree(null);  //调用,
}catch{
create();  //创建树的根节点方法
}

Answer 3

No need to go to such trouble.

void InvertTree(TreeNode* root)
{
        if(root)
        {
                TreeNode* tmp = root->left;
                root->left = root->right;
                root->right = tmp;
                InvertTree(root->left);
                InvertTree(root->right);
        }
}

Would this idea be more concise and clear?