我正在尝试创建一个用于淘汰赛的二叉树。该树由带有左右指针的TNode组成。
这是我提出的代码（下）;但是，CreateTree部分中的指针遇到了困难。
一旦创建了一个足够大的空树，我需要将Memo1.List上的名称添加到树的底部，这样我就可以将它们配对以进行匹配。
我该怎么做？
Type
TNodePtr = ^TNode;
TNode = Record
Data:String;
Left:TNodePtr;
Right:TNodePtr;
end;
Type
TTree = Class
Private
Root:TNodePtr;
Public
Function GetRoot:TNodePtr;
Constructor Create;
end;
var
MyTree:TTree;
function TTree.GetRoot: TNodePtr;
begin
Result:=Root;
end;
Constructor TTree.Create;
Var NewNode:TNodePtr;
Begin
New(NewNode);
NewNode^.Data:='Spam';
NewNode^.Left:=Nil;
NewNode^.Right:=Nil;
End;
Function Power(Base:integer;Exponent:integer):Integer; //Used for only positive powers in this program so does not need to handle negatives.
begin
If Base = 0 then
Power := 0
else If Exponent = 0 then
Power := 1
else //If Exponent > 0 then
Power:=Base*Power(Base, Exponent-1);
end;
Function DenToBinStr(Value:Integer):String;
Var iBinaryBit:integer;
sBinaryString:String;
Begin
While Value <> 0 do
begin
iBinaryBit:=Value mod 2;
sBinaryString:=sBinaryString+IntToStr(iBinaryBit);
Value:=Value div 2;
end;
Result:=sBinaryString;
end;
Procedure TForm1.CreateTree;
Var iRounds, iCurrentRound, iTreeLocation, iNodeCount, iMoreString, iAddedStringLength, iStringTree:Integer;
sBinary:String;
NewNode, ThisNode:TNodePtr;
begin
iRounds:=0;
While Power(2,iRounds) < Memo1.Lines.Count do {Calculates numbers of rounds by using whole powers of 2}
iRounds:=iRounds+1;
If iRounds > 0 then {Make sure there IS a round}
begin
For iCurrentRound:=1 to iRounds do {Select the round we are currently adding nodes to}
begin
iTreeLocation:=Power(2,iCurrentRound); {Works out the number of nodes on a line}
For iNodeCount:= 0 to iTreeLocation do {Selects the node we are currently working on}
begin
ThisNode:=MyTree.GetRoot;
sBinary:=DenToBinStr(iNodeCount); {Gets the tree traversal to that node from the root}
If Length(sBinary) < iCurrentRound then {Makes sure that the tree traversal is long enough, Fills spare spaces with Left because 0 decimal = 0 binary (we need 00 for 2nd round)}
begin
iMoreString:= iCurrentRound-Length(sBinary);
for iAddedStringLength := 0 to iMoreString do
sBinary:='0'+sBinary;
end;
iStringTree:=0; {Init iStringTree, iStringTree is the position along the binary string (alt the position down the tree)}
While iStringTree <= iCurrentRound-1 do {While we are not at the location to add nodes to, move our variable node down the tree}
begin
If sBinary[iStringTree]='0' then
ThisNode:=ThisNode^.Left
else If sBinary[iStringTree]='1' then
ThisNode:=ThisNode^.Right;
iStringTree:=iStringTree+1;
end;
New(NewNode); {Create a new node once we are in position}
NewNode^.Data:='Spam';
NewNode^.Left:=Nil;
NewNode^.Right:=Nil;
If sBinary[iCurrentRound]='0' then
ThisNode^.Left:=NewNode
else If sBinary[iCurrentRound]='1' then
ThisNode^.Right:=NewNode;
ThisNode.Data:='Spam';
Showmessage(ThisNode.Data);
end;
end;
end;
{1.2Add on byes}
{1.2.1Calculate No Of Byes and turn into count. Change each count into binary
equivalent then flip the bits}
//iByes:= Memo1.Lines.Count - Power(2,iRounds);
{1.2.2Add node where 0 is left and 1 is right}
{2THEN FILL TREE using If node.left and node.right does not exist then write
next name from list[q] q++}
{3THEN DISPLAY TREE}
end;
答案 0 :(得分：1)
考虑通过从树叶构建树来完全不同地构建树。如果您有一个节点队列，则可以从前面取两个节点，将它们与一个新节点连接在一起，然后将该新节点添加到队列的末尾。重复，直到你的节点用完为止，你将拥有一个锦标赛支架，你可以从尝试从根构建树时得到相同数量的轮次。
这是构建树和的代码，用备忘录中的名称填充树叶。
var
Nodes: TQueue;
Node: PNode;
s: string;
begin
Nodes := TQueue.Create;
try
// Build initial queue of leaf nodes
for s in Memo1.Lines do begin
New(Node);
Node.Data := s;
Node.Left := nil;
Node.Right := nil;
Nodes.Push(Node);
end;
// Link all the nodes
while Nodes.Count > 1 do begin
New(Node);
Node.Left := Nodes.Pop;
Node.Right := Nodes.Pop;
Nodes.Push(Node);
end;
Assert((Nodes.Count = 1) or (Memo1.Lines.Count = 0));
if Nodes.Empty then
Tree := TTree.Create
else
Tree := TTree.Create(Nodes.Pop);
finally
Nodes.Free;
end;
end;
该代码的优点在于，我们在任何时候都不知道或关心任何特定节点需要的级别。
如果竞争对手的数量不是2的幂，那么列表末尾的一些竞争者可能会获得“再见”轮次，并且他们将被安排在列表顶部玩赢家。上面的代码具有最少数量的节点。您的代码可能包含许多“垃圾”节点，这些节点并不代表锦标赛中的任何实际匹配。
树对象应该拥有它包含的节点，因此它应该有一个析构函数，如下所示：
destructor TTree.Destroy;
procedure FreeSubnodes(Node: PNode);
begin
if Assigned(Node.Left) then
FreeSubnodes(Node.Left);
if Assigned(Node.Right) then
FreeSubnodes(Node.Right);
Dispose(Node);
end;
begin
FreeSubnodes(Root);
inherited;
end;
你会注意到我也改变了树的构造函数的调用方式。如果树为空，则不需要任何节点。如果树不是空的，那么我们在创建它时会为它提供节点。
constructor TTree.Create(ARoot: PNode = nil);
begin
inherited;
Root := ARoot;
end;
如果您有机会复制树，那么您也需要复制其所有节点。如果不这样做，那么当你释放一棵树时，副本的根节点指针将突然变为无效。
constructor TTree.Copy(Other: TTree);
function CopyNode(Node: PNode): PNode;
begin
if Assigned(Node) then begin
New(Result);
Result.Data := Node.Data;
Result.Left := CopyNode(Node.Left);
Result.Right := CopyNode(Node.Right);
end else
Result := nil;
end;
begin
inherited;
Root := CopyNode(Other.Root);
end;
答案 1 :(得分：0)
我实际上设法重写了原始代码，使其单独工作。它似乎在这个时刻起作用。 这是我现在使用的程序。谢谢Rob，我会把你的答案设置为答案，因为看起来它对我来说会更好用，我会查看它以了解我能做什么，但为了不用不必要地使用其他代码，我现在就用我自己的代码
Procedure TForm1.CreateTree;
Var iRounds, iCurrentRound, iCurrentNode, iTraverseToNode:integer;
sBinary:String;
ThisNode, NewNode, NextNode:TNodePtr;
begin
iRounds:=0;
While Power(2,iRounds) < Memo1.Lines.Count do {Calculates numbers of rounds by using whole powers of 2}
iRounds:=iRounds+1;
If iRounds > 0 then
begin
for iCurrentRound:=1 to iRounds do
begin
for iCurrentNode:=0 to power(2,iCurrentRound)-1 do
begin
NextNode:=MyTree.GetRoot;
ThisNode:=NextNode;
New(NewNode);
NewNode.Data:='';
NewNode.Left:=Nil;
NewNode.Right:=Nil;
sBinary:=DenToBinStr(iCurrentNode);
if sBinary = '' then
sBinary:='0';
While length(sBinary)>iCurrentNode+1 do
begin
sBinary:='0'+sBinary;
end;
for iTraverseToNode:=1 to length(sBinary)-1 do
While NextNode <> nil do
begin
if sBinary[iTraverseToNode] = '0' then
begin
ThisNode:=NextNode;
NextNode:=NextNode.Left;
end
else if sBinary[iTraverseToNode] = '1' then
begin
ThisNode:=NextNode;
NextNode:=NextNode.Right;
end
end;
if sBinary[iCurrentNode+1] = '0' then
ThisNode^.Left:=NewNode
else if sBinary[iCurrentNode+1] = '1' then
ThisNode^.Right:=NewNode
else
Showmessage('TooFar');
break;
end;
end;
end;
end;
编辑：03/03/2010 我发现了一种更好，更简单的递归方式。
Procedure RecursiveTree(r:integer; ThisNode: TNodePtr);
Var NewNode:TNodePtr;
begin
If (NOT assigned(ThisNode.Left)) and (r<>0) then
begin
New(NewNode);
NewNode.Left:=Nil;
NewNode.Right:=Nil;
NewNode.Data:='';
ThisNode.Left:=NewNode;
RecursiveTree(r-1,ThisNode.Left);
end;
If (NOT assigned(ThisNode.Right)) and (r<>0) then
begin
New(NewNode);
NewNode.Left:=Nil;
NewNode.Right:=Nil;
NewNode.Data:='';
ThisNode.Right:=NewNode;
RecursiveTree(r-1,ThisNode.Right);
end;
end;