我在使用C＃设置子节点时遇到了一些问题。我正在尝试构建一个节点树，其中每个节点都包含一个int值，并且最多可以有一些等于它的值的子节点。
当我在一个寻找空（空）子节点的节点中进行迭代时，我的问题出现了，这样我就可以在该位置添加一个新节点。我可以找到并返回null节点，但是当我将新节点设置为它时，它会失去与父节点的连接。
因此，如果我添加1个节点，那么它会链接到我的头节点，但是如果我尝试添加第二个节点，它就不会成为头节点的子节点。我试图通过单元测试来构建它，所以这里是测试代码，显示头部确实没有显示新节点作为它的孩子（也通过visual studio调试器确认）：
[TestMethod]
public void addSecondNodeAsFirstChildToHead()
{
//arange
Problem3 p3 = new Problem3();
p3.addNode(2, p3._head);
Node expected = null;
Node expected2 = p3._head.children[0];
int count = 2;
//act
Node actual = p3.addNode(1, p3._head);
Node expected3 = p3._head.children[0];
//assert
Assert.AreNotEqual(expected, actual, "Node not added"); //pass
Assert.AreNotEqual(expected2, actual, "Node not added as first child"); //pass
Assert.AreEqual(expected3, actual, "Node not added as first child"); //FAILS HERE
Assert.AreEqual(count, p3.nodeCount, "Not added"); //pass
}
这是我的代码。
public class Node
{
public Node[] children;
public int data;
public Node(int value)
{
data = value;
children = new Node[value];
for(int i = 0; i < value; i++)
{
children[i] = null;
}
}
}
public class Problem3
{
public Node _head;
public int nodeCount;
public Problem3()
{
_head = null;
nodeCount = 0;
}
public Node addNode(int value, Node currentNode)
{
if(value < 1)
{
return null;
}
Node temp = new Node(value);
//check head
if (_head == null)
{
_head = temp;
nodeCount++;
return _head;
}
//start at Current Node
if (currentNode == null)
{
currentNode = temp;
nodeCount++;
return currentNode;
}
//find first empty child
Node emptyChild = findEmptyChild(currentNode);
emptyChild = temp;
nodeCount++;
return emptyChild;
}
public Node findEmptyChild(Node currentNode)
{
Node emptyChild = null;
//find first empty child of current node
for (int i = 0; i < currentNode.children.Length; i++)
{
if (currentNode.children[i] == null)
{
return currentNode.children[i];
}
}
//move to first child and check it's children for an empty
//**this causes values to always accumulate on left side of the tree
emptyChild = findEmptyChild(currentNode.children[0]);
return emptyChild;
}
我觉得问题是我试图将节点视为指针，就像在C ++中一样，但是它没有像我期望的那样工作。
答案 0 :(得分：2)
函数不可能将句柄（或指针）返回到尚不存在的东西。要么初始化函数内部不存在的值，要么提供足够的变量使其在函数外部初始化。
一种解决方案是将函数findEmptyChild重命名为initializeEmptyChild(Node currentNode, Node newNode)，向其添加一个Node参数（当调用它为temp值时） ，并在return之前的循环中初始化以前为空的Node，currentNode.children[i] = newNode。
另一个解决方案不是只返回一个Node而是返回两个值，一个父节点和一个找到空子项的索引Tuple<Node, int> findEmptyChild(Node currentNode)，而不是返回return currentNode.children[i]你做return new Tuple<Node, int>(currentNode, i)。调用该函数时，您将代码更改为
var parentAndIndex = findEmptyChild(currentNode);
parentAndIndex.Item1.children[parentAndIndex.Item2] = temp;
答案 1 :(得分：1)
查看代码的这一部分：
Node temp = new Node(value);
//...
Node emptyChild = findEmptyChild(currentNode);
emptyChild = temp;
您正在将emptyChild分配给新节点，这样做会“松散”与任何父节点的连接。你应该写这样的东西：
emptyChild.data = temp.data;
emptyChild.children = temp.children;
正如其他人所说，使用空检查的方法可以改进。您提到Node.data包含给定节点的子节点数，因此您可以简单地说，当您有Node.data == 0时，该节点应被视为空或空。例如，而不是：
rootNode.children[0] = null; // rootNode can have a lot of children
rootNode.children[1] = null;
//...
rootNode.children[0] = new Node(0);
rootNode.children[1] = new Node(0);
//...
此时您的代码看起来与此类似：
public class Node
{
public Node[] children;
public int data;
public Node(int value)
{
data = value;
children = new Node[value];
// Instead of "pointing" to null,
// create a new empty node for each child.
for (int i = 0; i < value; i++)
{
children[i] = new Node(0);
}
}
}
public class Problem3
{
public Node _head;
public int nodeCount;
public Problem3()
{
_head = null;
nodeCount = 0;
}
public Node addNode(int value, Node currentNode)
{
if (value < 1)
{
return null;
}
Node temp = new Node(value);
//check head
if (_head == null)
{
_head = temp;
nodeCount++;
return _head;
}
//start at Current Node
if (currentNode == null)
{
currentNode = temp;
nodeCount++;
return currentNode;
}
//find first empty child
Node emptyChild = findEmptyChild(currentNode);
if (emptyChild != null)
{
emptyChild.data = temp.data;
emptyChild.children = temp.children;
nodeCount++;
}
return emptyChild;
}
public Node findEmptyChild(Node currentNode)
{
// Null checking.
if (currentNode == null)
return null;
// If current node is empty, return it.
if (currentNode.data == 0)
return currentNode;
// If current node is non-empty, check its children.
// If no child is empty, null will be returned.
// You could change this method to check even the
// children of the children and so on...
return currentNode.children.FirstOrDefault(node => node.data == 0);
}
}
现在让我们看看测试部分（请参阅评论以获得澄清）：
[TestMethod]
public void addSecondNodeAsFirstChildToHead()
{
//arange
Problem3 p3 = new Problem3();
p3.addNode(2, p3._head); // Adding two empty nodes to _head, this means that now _head can
// contain two nodes, but for now they are empty (think of them as
// being "null", even if it's not true)
Node expected = null;
Node expected2 = p3._head.children[0]; // Should be the first of the empty nodes added before.
// Be careful: if you later change p3._head.children[0]
// values, expected2 will change too, because they are
// now pointing to the same object in memory
int count = 2;
//act
Node actual = p3.addNode(1, p3._head); // Now we add a non-empty node to _head, this means
// that we will have a total of two non-empty nodes:
// this one fresly added and _head (added before)
Node expected3 = p3._head.children[0]; // This was an empty node, but now should be non-empty
// because of the statement above. Now expected2 should
// be non-empty too.
//assert
Assert.AreNotEqual(expected, actual, "Node not added"); //pass
// This assert won't work anymore, because expected2, expected 3 and actual
// are now pointing at the same object in memory: p3._head.children[0].
// In your code, this assert was working because
// In order to make it work, you should replace this statement:
// Node expected2 = p3._head.children[0];
// with this one:
// Node expected2 = new Node(0); // Create an empty node.
// expected2.data = p3._head.children[0].data; // Copy data
// expected2.children = p3._head.children[0].children;
// This will make a copy of the node instead of changing its reference.
Assert.AreNotEqual(expected2, actual, "Node not added as first child");
// Now this will work.
Assert.AreEqual(expected3, actual, "Node not added as first child");
Assert.AreEqual(count, p3.nodeCount, "Not added"); //pass
}