创建二叉树的时间复杂度

Question

There is a method that builds a binary tree in which there are N full levels and on each level I there are two nodes whose information parts are equal to I. What is the time complexity depending on the number of levels of the tree built?

private SimpleTreeNode fromNDigitToNode(int x,int k) throws Exception {
        IndexWrapper iw = new IndexWrapper();
        T parentValue = readValue((k-x+1)+"",iw);
        SimpleTreeNode parentNode = new SimpleTreeNode(parentValue);
        SimpleTreeNode leftNode = new SimpleTreeNode(parentValue);
        SimpleTreeNode rightNode = new SimpleTreeNode(parentValue);
        if(x<=1)return parentNode;
        int z = --x;
            parentNode.left = fromNDigitToNode(z,k);
            parentNode.right = fromNDigitToNode(z,k);

        return parentNode;
    }

public void fromNDigit(String Ndigit) throws Exception{
        int digit = Integer.parseInt(Ndigit);

        IndexWrapper iw = new IndexWrapper();
        SimpleTreeNode root = fromNDigitToNode(digit,digit);

        this.root = root;
    }
}

Answer 1

I think your code is flawed. You create leftNode and rightNode, but you don't use them. Furthermore, both parentNode.left and parentNode.right are created with the same parameters. Is that correct?

I have no idea what your IndexWrapper is for and what readValue() does.

And the biggest problem: this never stops. You don't check (for instance) if x < 0. So you call it with x=1. You then recurse twice, calling it with x=0, then x=-1, and repeat forever.

So I'd say this is Order(forever).

Might I suggest you actually get your code to work, test it, and THEN figure out the complexity.