Recursively searching a Binary Tree for string values

Question

I've got a binary tree that should only hold unique string values. Prior to entering a new string (which is done by the user), I need to recursively check the tree to see if the string already exists. Here's the method I've come up with, but it's only finding certain values (the root and the left ones I believe). Any tips on how to fix this are appreciated!

public static TreeNode wordExists(TreeNode root, String strInput){
    if (root != null){

    if (strInput.equals(root.dataItem))
    {
        return root;
    }
    if (root.left != null){
        return wordExists (root.left, strInput);
    }
    if (root.right != null){
        return wordExists (root.right, strInput);
        }
    }
    return null;
}

jpm · Accepted Answer · 2012-11-21 03:55:13Z

2

When you navigate down each branch, you need to check the result before returning it. Otherwise, if the result is only in the right branch, but there are other, non-null values in the left, you'll just return null, since it isn't found in the left path.

So instead of

if (root.left != null) {
    return wordExists(root.left, strInput);
}
if (root.right != null) {
    return wordExists(root.right, strInput);
}

you might do something like

TreeNode result;
if ((result = wordExists(root.left, strInput)) != null) {
    return result;
}
return wordExists(root.right, strInput);

You can get away with the shortcut on the second recursion, since, if it fails, you'll just be returning null anyway.

answered Nov 21, 2012 at 3:55

jpm

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2 Comments

Jeff Over a year ago

Something along the lines of the following: if(root.left != null){TreeNode ret = wordExists(root.left); if(ret != null) return ret;}

jpm Over a year ago

Actually, since you check for null first thing in the method, you can just recurse without explicitly checking for the existence of root.left.

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