June 8, 2015
我最喜欢的一道算法题目, 二行代码.
编程序需要很强的逻辑思维, 多问几个为什么, 可不可以简化．想一想, 二行代码, 五分钟就可以搞定; 2015年网上大家热议的 Homebrew 的作者 Max Howell 面试
Google 挂掉的一题, 二叉树反转, 七行代码, 相比二行代码, 情有可原!
Problem: return the count of binary tree with only one child
想一想, 你要写几行, 六七行, 或小于十行?
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));
}
Github for source code:
https://github.com/jianminchen/leetcode-tree/blob/master/TreeDemo.cs