二进制搜索树中的递归toString（）方法。 这个的时间复杂度是多少？

Question

I'm a beginner in Java and looking for some help.

So I've made this binary tree in Java, and I'm supposed to implement a method which sorts all elements in order and convert them into a string. It's supposed to look like ex. "[1, 2, 3, 4]". I used the StringBuilder in order to do this.

My code for the method looks loke this:

/**
 * Converts all nodes in current tree to a string. The string consists of
 * all elements, in order.
 * Complexity: ?
 * 
 * @return string
 */
public String toString() {
    StringBuilder string = new StringBuilder("[");
    helpToString(root, string);
    string.append("]");
    return string.toString();
}

/**
 * Recursive help method for toString. 
 * 
 * @param node
 * @param string
 */
private void helpToString(Node<T> node, StringBuilder string) {
    if (node == null)
        return; // Tree is empty, so leave.

    if (node.left != null) {
        helpToString(node.left, string);
        string.append(", ");
    }

    string.append(node.data);

    if (node.right != null) {
        string.append(", ");
        helpToString(node.right, string);
    }
}

So my question is, how do I calculate the time complexity for this? Also, if there are any suggestions in how to make this method better, I would gladly appreciate it.

Answer 1

The easiest answer is: O(n) . You visit every node once and do one (a) amount of work. The calculation would go like

O(a*n)

and because we ignore constant factors, the simple answer would be

O(n)

But . One could also argue, your doing just a little bit more: you return null each time you visit a place where there is no leaf. This again is one (b) amount of work to be done.

Let's call those invisible leaves for a moment. Following this idea, every value in the tree is a node which has one or two invisible leafs .

So now, we have the following to do (for each node):

a       | if a node has two child nodes
a + b   | if a node has one child node
a + 2b  | if a node has no child node.

for a worst case binary tree (totally unbalanced), we have (n-1) nodes with one child node and one node with no child node:

  "1"
    \
    "2"
      \
      "3"
        \
        "4"

And so, the calculation is

     (n-1)*(a+b) + b
 <=> an + bn - a - b + b
 <=> n(a+b) + b

  => O(an + bn)  // we ignore `+ b` because we always look at really big n's

Fortunately, even in a worst case time scenario, complexity is still linear. Only the constant is higher than in the easy answer. In most cases - usually when Big-O is needed to compare algorithms - we even ignore the factor and we're happy to say, that the algorithms time complexity is linear (O(n)).

Answer 2

The time complexity is O(n) since you are visiting every node once. You cannot do any better than that in order walk the tree.

Answer 3

时间复杂度在树中的节点数中呈线性关系：您正在访问每个节点一次。