在BST中计算少于X的节点数

Question

I have a code that is giving me these problems. Can you give me some hint or guide about the mistake I am making, or any change that I should make? The errors are:

BSTNode'<'Golfer'>' cannot be converted to BinarySearchTree'<'Golfer'>'.

public int countLess (BinarySearchTree <Golfer> tree, int value) {
BSTNode<Golfer> node = tree.node;

if (node == null) 
return 0;

int left = countLess(node.getLeft(), value);
int right  = countRight(node.getRight(), value);
return (node.getInfo() > maxValue ? 1:0) + countLeft + countRight;
}

Answer 1

I think it should be something like this as i am guessing node.getLeft() actually gives you a node in BST not the complete Left Subtree.

public int countLess (BSTNode <Golfer> node, int value) {
    if (node == null) 
         return 0;
    int left = countLess(node.getLeft(), value);
    int right  = countLess(node.getRight(), value);
    return (node.getInfo() > maxValue ? 1:0) + left + right;
}

Hope this solves your issue. I can provide a more correct solution if you can share Implementation Of BinarySearchTree and BSTNode classes implementation.

Answer 2

I think you should do an inorder traversal of the BST . The inorder traversal of a BST always gives you element in ascending order. Just keep a count variable while doing inorder traversal (and keep incrementing it for each node visited), and as soon the value of any node becomes more than X, just 'break'.

The final value in your count variable will be the answer.

BST: Binary Search Tree