我正在尝试提高应用中的效果。我已经以调用树的形式获得了性能信息，具有以下节点类：
public class Node
{
public string Name; // method name
public decimal Time; // time spent in method
public List<Node> Children;
}
我想打印出树，这样我就能看到节点之间的线条 - 就像在this question中一样。我可以在C＃中使用什么算法来做到这一点？
编辑：显然我需要使用递归 - 但我的尝试继续将行放在错误的位置。我要求的是一种特定的算法，它将以一种很好的方式打印树 - 有关何时打印垂直线以及何时打印水平线的详细信息。
编辑：仅使用字符串的副本来缩进节点是不够的。我不是在寻找
A
|-B
|-|-C
|-|-D
|-|-|-E
|-F
|-|-G
必须是
A
+-B
| +-C
| +-D
| +-E
+-F
+-G
或类似的东西，只要树结构可见即可。请注意，C和D的缩写与G不同 - 我不能只使用重复的字符串来缩进节点。
答案 0 :(得分：75)
诀窍是传递一个字符串作为缩进并特别处理最后一个孩子：
class Node
{
public void PrintPretty(string indent, bool last)
{
Console.Write(indent);
if (last)
{
Console.Write("\\-");
indent += " ";
}
else
{
Console.Write("|-");
indent += "| ";
}
Console.WriteLine(Name);
for (int i = 0; i < Children.Count; i++)
Children[i].PrintPretty(indent, i == Children.Count - 1);
}
}
如果这样调用：
root.PrintPretty("", true);
将以此样式输出：
\-root
\-child
|-child
\-child
|-child
|-child
\-child
|-child
|-child
| |-child
| \-child
| |-child
| |-child
| |-child
| \-child
| \-child
| \-child
\-child
|-child
|-child
|-child
| \-child
\-child
\-child
答案 1 :(得分：22)
当您深入树中时，您需要跟踪已修改的缩进字符串。为避免添加额外的|个字符，您还需要知道该节点是否是该集合中的最后一个子节点。
public static void PrintTree(Node tree, String indent, Bool last)
{
Console.Write(indent + "+- " + tree.Name);
indent += last ? " " : "| ";
for (int i == 0; i < tree.Children.Count; i++)
{
PrintTree(tree.Children[i], indent, i == tree.Children.Count - 1);
}
}
当这样调用时：
PrintTree(node, "", true)
它将输出如下文字：
+- root
+- branch-A
| +- sibling-X
| | +- grandchild-A
| | +- grandchild-B
| +- sibling-Y
| | +- grandchild-C
| | +- grandchild-D
| +- sibling-Z
| +- grandchild-E
| +- grandchild-F
+- branch-B
+- sibling-J
+- sibling-K
如果您碰巧有一个非常深度树并且您的调用堆栈大小有限，您可以改为执行静态的非递归树遍历来输出相同的结果：
public static void PrintTree(Node tree)
{
List<Node> firstStack = new List<Node>();
firstStack.Add(tree);
List<List<Node>> childListStack = new List<List<Node>>();
childListStack.Add(firstStack);
while (childListStack.Count > 0)
{
List<Node> childStack = childListStack[childListStack.Count - 1];
if (childStack.Count == 0)
{
childListStack.RemoveAt(childListStack.Count - 1);
}
else
{
tree = childStack[0];
childStack.RemoveAt(0);
string indent = "";
for (int i = 0; i < childListStack.Count - 1; i++)
{
indent += (childListStack[i].Count > 0) ? "| " : " ";
}
Console.WriteLine(indent + "+- " + tree.Name);
if (tree.Children.Count > 0)
{
childListStack.Add(new List<Node>(tree.Children));
}
}
}
}
答案 2 :(得分：9)
创建PrintNode方法并使用递归：
class Node
{
public string Name;
public decimal Time;
public List<Node> Children = new List<Node>();
public void PrintNode(string prefix)
{
Console.WriteLine("{0} + {1} : {2}", prefix, this.Name, this.Time);
foreach (Node n in Children)
if (Children.IndexOf(n) == Children.Count - 1)
n.PrintNode(prefix + " ");
else
n.PrintNode(prefix + " |");
}
}
然后打印整个树只执行：
topNode.PrintNode("");
在我的例子中，它会给我们类似的东西：
+ top : 123
| + Node 1 : 29
| | + subnode 0 : 90
| | + sdhasj : 232
| | + subnode 1 : 38
| | + subnode 2 : 49
| | + subnode 8 : 39
| + subnode 9 : 47
+ Node 2 : 51
| + subnode 0 : 89
| + sdhasj : 232
| + subnode 1 : 33
+ subnode 3 : 57
答案 3 :(得分：6)
以下是@Will对（当前接受的）答案的修改。变化是：
作为伪代码提供，以便在C ++之外更容易使用：
def printHierarchy( item, indent )
kids = findChildren(item) # get an iterable collection
labl = label(item) # the printed version of the item
last = isLastSibling(item) # is this the last child of its parent?
root = isRoot(item) # is this the very first item in the tree?
if root then
print( labl )
else
# Unicode char U+2514 or U+251C followed by U+2574
print( indent + (last ? '└╴' : '├╴') + labl )
if last and isEmpty(kids) then
# add a blank line after the last child
print( indent )
end
# Space or U+2502 followed by space
indent = indent + (last ? ' ' : '│ ')
end
foreach child in kids do
printHierarchy( child, indent )
end
end
printHierarchy( root, "" )
示例结果：
Body
├╴PaintBlack
├╴CarPaint
├╴Black_Material
├╴PaintBlue
├╴Logo
│ └╴Image
│
├╴Chrome
├╴Plastic
├╴Aluminum
│ └╴Image
│
└╴FabricDark
答案 4 :(得分：5)
我使用以下方法打印BST
private void print(Node root, String prefix) {
if (root == null) {
System.out.println(prefix + "+- <null>");
return;
}
System.out.println(prefix + "+- " + root);
print(root.left, prefix + "| ");
print(root.right, prefix + "| ");
}
以下是输出。
+- 43(l:0, d:1)
| +- 32(l:1, d:3)
| | +- 10(l:2, d:0)
| | | +- <null>
| | | +- <null>
| | +- 40(l:2, d:2)
| | | +- <null>
| | | +- 41(l:3, d:0)
| | | | +- <null>
| | | | +- <null>
| +- 75(l:1, d:5)
| | +- 60(l:2, d:1)
| | | +- <null>
| | | +- 73(l:3, d:0)
| | | | +- <null>
| | | | +- <null>
| | +- 100(l:2, d:4)
| | | +- 80(l:3, d:3)
| | | | +- 79(l:4, d:2)
| | | | | +- 78(l:5, d:1)
| | | | | | +- 76(l:6, d:0)
| | | | | | | +- <null>
| | | | | | | +- <null>
| | | | | | +- <null>
| | | | | +- <null>
| | | | +- <null>
| | | +- <null>
答案 5 :(得分：1)
这是Joshua Stachowski的答案的通用版本。关于Joshua Stachowski的回答是好处，它不需要实际的节点类来实现任何额外的方法，它看起来也不错。
我使他的解决方案通用，可以在不修改代码的情况下用于任何类型。
PrintTree(rootNode, x => x.ToString(), x => x.Children);
用法
<DataGrid x:Name="TheList" Grid.Row="1" ItemsSource="{Binding People}" SelectionMode="Extended" SelectionUnit="FullRow" AutoGenerateColumns="False"
<DataGrid.Columns>
<DataGridCheckBoxColumn Binding="{Binding IsSelected}">
<DataGridCheckBoxColumn.HeaderTemplate>
<DataTemplate>
<CheckBox IsChecked="{Binding Path=DataContext.AllSelected, RelativeSource={RelativeSource AncestorType=DataGrid}}"></CheckBox>
</DataTemplate>
</DataGridCheckBoxColumn.HeaderTemplate>
</DataGridCheckBoxColumn>
<DataGridTextColumn Binding="{Binding GivenName}" Header="Given Name" />
...
<DataGridTextColumn Binding="{Binding BranchName}" Header="Branch" />
</DataGrid.Columns>
</DataGrid>
答案 6 :(得分：0)
没有使用递归的完全可选性的最佳方法是` https://github.com/tigranv/Useful_Examples/tree/master/Directory%20Tree
public static void DirectoryTree(string fullPath)
{
string[] directories = fullPath.Split('\\');
string subPath = "";
int cursorUp = 0;
int cursorLeft = 0;
for (int i = 0; i < directories.Length-1; i++)
{
subPath += directories[i] + @"\";
DirectoryInfo directory = new DirectoryInfo(subPath);
var files = directory.GetFiles().Where(f => !f.Attributes.HasFlag(FileAttributes.Hidden)).Select(f => f.Name).ToArray();
var folders = directory.GetDirectories().Where(f => !f.Attributes.HasFlag(FileAttributes.Hidden)).Select(f => f.Name).ToArray();
int longestFolder = folders.Length != 0 ? (folders).Where(s => s.Length == folders.Max(m => m.Length)).First().Length:0;
int longestFle = files.Length != 0? (files).Where(s => s.Length == files.Max(m => m.Length)).First().Length : 0;
int longestName =3 + (longestFolder <= longestFle ? longestFle:longestFolder)<=25? (longestFolder <= longestFle ? longestFle : longestFolder) : 26;
int j = 0;
for (int k = 0; k < folders.Length; k++)
{
folders[k] = folders[k].Length <= 25 ? folders[k] : (folders[k].Substring(0, 22) + "...");
if (folders[k] != directories[i + 1])
{
Console.SetCursorPosition(cursorLeft, cursorUp + j);
Console.WriteLine("+" + folders[k]);
j++;
}
else
{
if (i != directories.Length - 2)
{
Console.SetCursorPosition(cursorLeft, cursorUp + j);
Console.WriteLine("-" + folders[k] + new string('-', longestName - directories[i + 1].Length) + "--\u261B");
j++;
}
else
{
Console.ForegroundColor = ConsoleColor.Red;
Console.SetCursorPosition(cursorLeft, cursorUp + j);
Console.WriteLine("***"+ folders[k] + "***");
Console.ForegroundColor = ConsoleColor.Gray;
j++;
}
}
}
for(int k = 0; k < files.Length; k++)
{
files[k] = files[k].Length <= 25 ? files[k] : (files[k].Substring(0, 22) + "...");
Console.SetCursorPosition(cursorLeft, cursorUp + j);
Console.WriteLine("+" + files[k]);
j++;
}
cursorUp += Array.IndexOf(folders, directories[i+1]) + 1;
cursorLeft += longestName+3;
}
}
答案 7 :(得分：0)
C代码：
void printVLine(wchar_t token, unsigned short height, unsigned short y, unsigned short x);
const static wchar_t TREE_VLINE = L'┃';
const static wchar_t TREE_INBRANCH[] = L"┣╾⟶ ";
const static wchar_t TREE_OUTBRANCH[] = L"┗╾⟶ ";
typedef void (*Printer)(void * whateverYouWant);
const static unsigned int INBRANCH_SIZE = sizeof(TREE_INBRANCH) / sizeof(TREE_INBRANCH[0]);
const static unsigned int OUTBRANCH_SIZE = sizeof(TREE_OUTBRANCH) / sizeof(TREE_OUTBRANCH[0]);
size_t Tree_printFancy(Tree * self, int y, int x, Printer print){
if (self == NULL) return 0L;
//
size_t descendants = y;
move(y, x);
print(Tree_at(self));
if (!Tree_isLeaf(self)){ // in order not to experience unsigned value overflow in while()
move(++y, x);
size_t i = 0;
while(i < Tree_childrenSize(self) - 1){
wprintf(TREE_INBRANCH);
size_t curChildren = Tree_printFancy(
Tree_childAt(self, i), y, x + INBRANCH_SIZE, print
);
printVLine(TREE_VLINE, curChildren , y + 1, x);
move((y += curChildren), x);
++i;
}
wprintf(TREE_OUTBRANCH);
y += Tree_printFancy( // printing outermost child
Tree_childAt(self, i), y, x + OUTBRANCH_SIZE, print
) - 1;
}
return y - descendants + 1;
}
适用于控制台打印。 函数move（y，x）将光标移动到屏幕上的（y，x）位置。 最好的部分是，您可以通过更改变量来更改输出样式 TREE_VLINE，TREE_INBRANCH，TREE_OUTBRANCH，最后两个字符串的长度无关紧要。您可以通过传递打印机功能指针来打印您喜欢的任何内容，该指针将打印当前树节点的值。 输出看起来像this
