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Thread: binarysearchtree traversing

  1. #1
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    Unhappy binarysearchtree traversing

    Hi, I'm trying to figure out how to do this contains method. everything that's in the method is my code. My thinking is: start at root and go in Inorder traversal and use .equals(). Should I even do that? is there a way I can compare the data, since BST will have lower value to left and higher in right child? I'm lost.
    /**
    Searches for a given element in the binary search tree
    @param someElement element to be searched
    @return true - if someElement is found in the tree; false otherwise
    */
    // Complexity: O(h) - where h is the height of the tree. In the worst case it could be O(n).  But on average
    // we can expect a complexity of O(log n)
    public boolean contains( E someElement) {
     
    	Node node1 = root;
    	if(root == someElement)
    		return true;
     
     
    	if(node1.left != null){
    		boolean result = someElement.equals(node1.left);
    		if(result == false){
    			contains((E) node1);
     
    		}
     
    	}
    }


  2. #2
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    Default Re: binarysearchtree traversing

    Please edit your post and wrap your code with code tags:
    [code=java]
    YOUR CODE HERE
    [/code]
    to get highlighting and preserve formatting.
    If you don't understand my answer, don't ignore it, ask a question.

  3. #3
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    Default Re: binarysearchtree traversing

    done! thanks

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    Default Re: binarysearchtree traversing

    Is someElement a Node or contained in a Node?
    If you don't understand my answer, don't ignore it, ask a question.

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    Default Re: binarysearchtree traversing

    Quote Originally Posted by Norm View Post
    Is someElement a Node or contained in a Node?
    I'm pretty sure it's contained in a node.

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    Default Re: binarysearchtree traversing

    Your best bet might be to use a recursive call on both child nodes.

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    Default Re: binarysearchtree traversing

    ok so this is what i have so far. does it make sense??
    	Node node1 = root;
    	boolean result = false;
     
    	if(root == someElement)
    		return true;
    	//go through left
    	if(node1.left != null){
    		result = someElement.equals(node1.left);
    		if(result == false){
    			contains((E) node1.left);
    		}
    	}
    	//go through right
    	if(node1.right != null){
    		result = someElement.equals(node1.right);
    		if(result == false){
    			contains((E) node1.right);
    		}
    return result;
    	}

  8. #8
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    Default Re: binarysearchtree traversing

    does it make sense??
    Not if someElement is not a Node. The code needs to compare against the contents of a Node.

    Can you make a complete program for testing this method? Build a simple tree and call the method. Add some println() statements for debugging and execute it to see what happens.
    If you don't understand my answer, don't ignore it, ask a question.

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    Default Re: binarysearchtree traversing

    Quote Originally Posted by Norm View Post
    Not if someElement is not a Node. The code needs to compare against the contents of a Node.

    Can you make a complete program for testing this method? Build a simple tree and call the method. Add some println() statements for debugging and execute it to see what happens.
    if I cast node1 as <E> will it make a difference?

  10. #10
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    Default Re: binarysearchtree traversing

    Casting will not make it a Node. You need to look at the API for the Node class and use one of its methods to access its contents which should be the same type as someElement.
    If you don't understand my answer, don't ignore it, ask a question.

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