/**
*
* Shortest Solution:
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, 面试问你: 优化代码;
* 这时你可以推敲, 哪句代码会有错误, 可能错在哪里.
* 提示: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));
}
/**
* More readable code.
*
*/
public static int countOneChildNode2(Node node)
{
if (node == null)
return 0;
bool having_l_child = node.left !=null;
bool having_r_child = node.right != null;
// either left child or right child existing,
// one true, another is false;
// two of them are true - two children
// none of them is true - no child
bool oneChildOnly = having_l_child != having_r_child;
return (oneChildOnly ? 1 : 0)
+ countOneChildNode2(node.left)
+ countOneChildNode2(node.right);
}
static void Main(string[] args)
{
/* Updated on July 9, 2015
* Test case:
* 1
* 2 3
* 4 5 6
* Result: one child only - value is 1, node is 3
*/
Node n1 = new Node(1);
n1.left = new Node(2);
n1.right = new Node(3);
n1.left.left = new Node(4);
n1.left.right = new Node(5);
n1.right.left = new Node(6);
int r1 = countOneChildNode1(n1);
int r2 = countOneChildNode(n1);
/*
* Test case 2:
* 1
* 2
* 3
* result: 2
*/
Node t1 = new Node(1);
t1.left = new Node(2);
t1.left.right = new Node(3);
int t3 = countOneChildNode(t1);
int val4 = countOneChildNode(t1);
}
/**
Latest update:
* The solution needs some improvement.
* See how good you are to improve it!
* julia comment:
* 1. Since first line is the discussion of “node==null”,
* so, it is safe to call the function with null pointer;
* 2. A node may have 1 child, 2 children or no children, 3 cases;
* 3. We need to find out 1 child only;
* need to simplify the checking
* 4. The code is too much to read:
* 4.1. function call 4 times
* 4.2 left!=null twice, right!=null twice, "+ 1" twice
* 4.3 if cases - how to merge some cases - minimum 2 cases
* Also, 0 repeats twice.
* Overall, there is too much places which may go wrong.
*
*/
public static int countOneChildNode1(Node node)
{
if (node == null)
return 0;
if (node.left != null && node.right != null) // two children
{
return countOneChildNode1(node.left) + countOneChildNode1(node.right);
}
else if (node.left != null) // left child only
{
return countOneChildNode1(node.left) + 1;
}
else if (node.right != null) // right child only
{
return countOneChildNode1(node.right) + 1;
}
else // no left child, no right child
return 0;
}
