一道算法题目, 二行代码, Binary Tree

Solution:  two lines code:

 /**

* 

* Great arguments:

1. Since first line is the discussion of “node==null”, there is no need to check node!=null before the function countOneChildNode call.

2. How to express only one child?

case 1: left child is not null; in other words, there is a left child: node.left!=null

case 2: right child is not null; node.right!=null)

case 3: node has two child

(node.left!=null) && (node.right!=null)

case 4: node has only one child (A: left child only, B: right child only, one true, one false; left child existed != right child existed; cannot be both false or both true)

(node.left!=null)!=(node.right!=null)

case 5: at least one child (one child or two child)

(node.left!=null) || (node.right!=null)

这道题非常好, 通过这道题, 你可以如何简化代码; 如果有一个出色的程序员, 有很强的逻辑思维能力, 想到只有一个孩子, 可以表示为一句话:　 (node.left!=null)!=(node.right!=null)

*/

public static int countOneChildNode(Node node)

{

if(node==null) return 0;

return (((node.left!=null)!=(node.right!=null)?1:0)+countOneChildNode(node.left)+countOneChildNode(node.right));

}