Java 打印二叉搜索树中序遍历
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Printing Binary Search Tree in-order Traversal
提问by user3415930
I've created a program that stores integers from user input in a binary search tree and I have recursive functions for pre, post and in-order traversals that work fine. What I'm trying to do is traverse the tree in-order and at each node I want to print the number that is stored there and the number in the node to the left and right of it or if the node is a leaf node. Assuming the user enters the integers 1,4,11 and 12 I want my output to look like:
我创建了一个程序，该程序将用户输入的整数存储在二叉搜索树中，并且我有递归函数用于 pre、post 和 in-order 遍历，可以正常工作。我想要做的是按顺序遍历树，在每个节点上，我想打印存储在那里的数字以及节点中左侧和右侧的数字，或者该节点是否是叶节点。假设用户输入整数 1、4、11 和 12，我希望我的输出如下所示：
1: Right Subtree: 12
1：右子树：12
4: Right Subtree: 11
4：右子树：11
11: Leaf node
11：叶节点
12: Left Subtree: 4 etc
12：左子树：4等
here is the code I am using for the function, when I run the program I am getting a null pointer exception.
这是我用于该函数的代码，当我运行该程序时，出现空指针异常。
public synchronized void inorderTraversal()
{ inorderHelper( root ); }
// Recursive method to perform inorder traversal
// 执行中序遍历的递归方法
private void inorderHelper( TreeNode node )
{
if ( node == null )
return;
inorderHelper( node.left );
System.out.print( node.data + ": Left Subtree " + node.left.data +": Right Subtree " + node.right.data);
inorderHelper( node.right );
}
回答by Anthony Accioly
You should only print node.data, the recursion will take care of printing the left and right trees in-order.
您应该只打印node.data，递归将负责按顺序打印左右树。
回答by Hypothetical inthe Clavicle
Chances are, your recursion is taking you to the bottom level of your tree (your leaves) and when you attempt to call
很有可能，您的递归会将您带到树的底层（您的叶子），并且当您尝试调用
node.left.data
it's a null => NullPointerException.
它是一个 null => NullPointerException。
As others are saying, just let your recursion take care of the work.
正如其他人所说，让你的递归来处理工作。
private void inorderHelper( TreeNode node )
{
if ( node == null )
return;
inorderHelper( node.left );
System.out.print( "Node data: " + node.data);
inorderHelper( node.right );
}
