您是否注意到您的insert() 方法无法处理“key-collision”的情况（由于最后一个if 缺少else 分支）。这可用于检测是否应插入已经在树中的密钥。事实上，这种重复的插入会被简单地忽略（恕我直言，这不是最糟糕的行为）。
在下面的示例中，我决定使用不同的选项：让insert() 返回一个布尔值，报告插入成功。
很遗憾，您没有提供MCVE。
因此，我用自己的代码填补了空白，其中（为了我个人的快乐）我涉及模板。希望这不会带来太多混乱。
但是，在处理了您的示例数据之后，我得出的结论是，您的代码可能是正确的——可能不是您所期望的。最终，你有一个错误的期望。或者，您正确地解决了错误问题。我多次重新阅读了您的问题，但没有清楚地了解这一点...
我使用您的最后一个示例文件作为输入（构建二叉搜索树）并让我的应用程序输出前序和中序：
预购输出对应于您的最后一个文件。
按顺序输出对应于您的第三个文件。
inorder 输出根据树的定义顺序提供排序输出 - 在这种情况下，莫尔斯序列的字典顺序（其中 - 位于 . 之前，因为它们各自的 ASCII 值）。
但这也不会按字母顺序打印出用于预购的摩尔斯电码的树。
嗯。如果您希望按字母顺序排列，您的意思是指考虑字母数字字符的顺序（莫尔斯电码映射到该顺序）？
如果是这样，这种树很难做到这一点（因为我看不出摩尔斯电码的可能顺序如何对应字母数字的可能顺序）。 IE。为了实现“按相关字母数字排序的莫尔斯电码”，显而易见的（并且仅限恕我直言）方法是为反向映射构建一棵树。您可以轻松地构建另一棵树（例如从第一棵树开始），其中分配的字母数字值用作键。 （实际上，您已经为字母数字创建了一个二叉搜索树。）
这让我有些困惑。可能是，我错过了一些东西，并没有得到你的实际问题......
但是，下面是我摆弄的结果：
// forward template declaration:
template <typename KEY, typename VALUE, typename COMP>
class BSTreeT;
/* provides a class template for a node in a binary search tree.
*
* KEY ... C++ type of the key values of nodes
* VALUE ... C++ type of the other values of nodes
*/
template <typename KEY, typename VALUE>
class BSTreeNodeT {
/* This friend shall ensure that the corresponding
* BSTreeT template class may access private _pLeft and _pRight.
*/
template <typename KEY_, typename VALUE_, typename COMP_>
friend class BSTreeT;
public:
// the key value of node (used to define an order)
const KEY key;
// other values of node
VALUE value;
private:
// pointers to left and right child nodes
BSTreeNodeT *_pLeft, *_pRight;
public:
// constructor.
BSTreeNodeT(const KEY &key, const VALUE &value):
key(key), value(value), _pLeft(nullptr), _pRight(nullptr)
{ }
// destructor.
~BSTreeNodeT() { delete _pLeft; delete _pRight; }
// disabled:
BSTreeNodeT(const BSTreeNodeT&) = delete;
BSTreeNodeT& operator=(const BSTreeNodeT&) = delete;
public:
// returns pointer to left child node (or nullptr if there is none).
const BSTreeNodeT* getLeft() const { return _pLeft; }
// returns pointer to right child node (or nullptr if there is none).
const BSTreeNodeT* getRight() const { return _pRight; }
};
/* provides a less functor which simply wraps operator < for a certain
* type
*
* VALUE ... C++ type of value to less-compare
*
* Note:
* This is actually, what std::less provides.
* I involved this functor for illustration.
*/
template <typename VALUE>
struct lessFunc {
bool operator()(const VALUE &value1, const VALUE &value2) const
{
return value1 < value2;
}
};
/* provides a class template for a binary search tree.
*
* KEY ... C++ type of the key values of nodes
* VALUE ... C++ type of the other values of nodes
* COMP ... C++ type of the less comparator
* to define an order of tree nodes
*/
template <typename KEY, typename VALUE, typename COMP = lessFunc<KEY> >
class BSTreeT {
public:
const COMP ∁
private:
BSTreeNodeT<KEY, VALUE> *_pRoot;
public:
/* constructor.
*
* comp ... a less comparator to define order of nodes
*/
explicit BSTreeT(const COMP &comp = COMP()):
comp(comp), _pRoot(nullptr)
{ }
// destructor.
~BSTreeT() { delete _pRoot; }
// disabled:
BSTreeT(const BSTreeT&) = delete;
BSTreeT& operator=(const BSTreeT&) = delete;
public:
/* inserts a node.
*
* key ... the key value of node
* value ... the other value of node
* return: true ... key/value inserted
* false ... Error! Possible reasons:
* - duplicated key
* - allocation of node failed.
*/
bool insert(const KEY &key, const VALUE &value)
{
return insert(_pRoot, key, value);
}
/* provides a functor-like type which is applied to every node
* in traverse().
*
* If an instance of this class is provided the traverse() does nothing
* else than the pure traversal.
*/
struct Apply {
// pre-order access to node
virtual void preNode(BSTreeNodeT<KEY, VALUE> &node) { }
// in-order access to node
virtual void inNode(BSTreeNodeT<KEY, VALUE> &node) { }
// post-order access to node
virtual void postNode(BSTreeNodeT<KEY, VALUE> &node) { }
};
/* traverses the tree and applies the provided object to every node.
*
* apply ... the action object applied to every node
*/
void traverse(Apply &apply)
{
if (_pRoot) traverse(_pRoot, apply);
}
private:
// inserts a node.
bool insert(
BSTreeNodeT<KEY, VALUE> *&pTree, const KEY &key, const VALUE &value)
{ /* Every if-branch ends with return.
* Thus, no explict else is needed.
*/
if (!pTree) { /* (!pTree) ... (pTree == nullptr) */
return !!(pTree = new BSTreeNodeT<KEY, VALUE>(key, value));
}
if (comp(key, pTree->key)) return insert(pTree->_pLeft, key, value);
if (comp(pTree->key, key)) return insert(pTree->_pRight, key, value);
return false;
}
// traverses the tree.
void traverse(BSTreeNodeT<KEY, VALUE> *pTree, Apply &apply)
{
apply.preNode(*pTree);
if (pTree->_pLeft) traverse(pTree->_pLeft, apply);
apply.inNode(*pTree);
if (pTree->_pRight) traverse(pTree->_pRight, apply);
apply.postNode(*pTree);
}
};
// sample code:
#include <ctype.h>
#include <iostream>
#include <string>
using namespace std;
// template instances (for convenience)
typedef BSTreeNodeT<string, string> BSTNode;
typedef BSTreeT<string, string> BST;
/* a helper function to split a string into tow at the first occurence of
* (a sequence of) whitespaces.
*
* line ... input
* first ... returns first sub-string (might become empty)
* second ... returns second sub-string (might become empty)
*/
void split(const string &line, string &first, string &second)
{
size_t i0 = 0, n = line.length(), i;
for (i = i0; i < n && !isspace(line[i]); ++i);
first = line.substr(i0, i - i0);
for (i0 = i; i0 < n && isspace(line[i0]); ++i0);
for (i = i0; i < n && !isspace(line[i]); ++i);
second = line.substr(i0, i - i0);
}
/* a derived tree-traversal action
* for graphical (i.e. ASCII-art) output of tree
*/
struct PrintGraph: public BST::Apply {
string indent;
PrintGraph(): indent(" ") { }
virtual void preNode(BSTNode &node)
{
indent.pop_back(); char c = indent.back(); indent.pop_back();
cout << indent << "+-"
<< (node.getLeft() || node.getRight() ? '+' : '-')
<< "- ["
<< node.key << ": " << node.value
<< ']' << endl;
indent += c; indent += ' ';
indent += node.getRight() ? "| " : " ";
}
virtual void inNode(BSTNode &node)
{
indent.pop_back(); indent.pop_back();
indent += " ";
}
virtual void postNode(BSTNode &node)
{
indent.pop_back(); indent.pop_back();
}
};
/* a derived tree-traversal action
* for pre-order output of nodes
*/
struct PrintPreOrder: public BST::Apply {
virtual void preNode(BSTNode &node)
{
cout << node.key << ": " << node.value << endl;
}
};
/* a derived tree-traversal action
* for in-order output of nodes
*/
struct PrintInOrder: public BST::Apply {
virtual void inNode(BSTNode &node)
{
cout << node.key << ": " << node.value << endl;
}
};
/* a derived tree-traversal action
* to fill another tree with key and value of nodes swapped
*/
struct FillRevTree: public BST::Apply {
BST &tree; // destination tree to fill
FillRevTree(BST &tree): tree(tree) { }
virtual void preNode(BSTNode &node)
{
tree.insert(node.value, node.key);
}
};
// main function
int main()
{
BST tree;
// read tree from input
cout << "Read contents from input:" << endl;
for (string line; getline(cin, line);) {
string key, value; split(line, key, value);
if (!tree.insert(key, value)) {
cerr << "Error! Couldn't store the line:" << endl
<< "->" << line << endl;
}
}
cout << "End of input." << endl
<< endl;
// show tree
cout << "The tree:" << endl;
{ PrintGraph print; tree.traverse(print); }
cout << endl;
// print tree by pre-order traversal
cout << "Pre-Order Output:" << endl;
{ PrintPreOrder print; tree.traverse(print); }
cout << endl;
// print tree by in-order traversal
cout << "In-Order Output:" << endl;
{ PrintInOrder print; tree.traverse(print); }
cout << endl;
// re-built tree with keys and values swapped
BST treeRev;
{ FillRevTree fill(treeRev); tree.traverse(fill); }
// show reverse tree
cout << "The Rev. Tree:" << endl;
{ PrintGraph print; treeRev.traverse(print); }
cout << endl;
// print tree by in-order traversal
cout << "In-Order Output of Rev. Tree:" << endl;
{ PrintInOrder print; treeRev.traverse(print); }
cout << endl;
// done
cout << "That's it." << endl;
return 0;
}
编译运行：
$ g++ -std=c++11 -o binary-search-tree binary-search-tree.cc
$ ./binary-search-tree <morse.txt
Read contents from input:
End of input.
The tree:
+-+- [-..: D]
+-+- [--.-: Q]
| +-+- [-----: 0]
| | +-+- [--: M]
| | | +--- [-: T]
| | | +--- [---: O]
| | +-+- [---..: 8]
| | +--- [----.: 9]
| | +--- [--.: G]
| +-+- [-.: N]
| +-+- [--..--: ,]
| | +--- [--..: Z]
| | +--- [--...: 7]
| +-+- [-.--: Y]
| +--- [-.-: K]
| +--- [-.-.: C]
+-+- [..: I]
+-+- [.----: 1]
| +-+- [.: E]
| | +-+- [-...: B]
| | | +--- [-..-: X]
| | | +--- [-....: 6]
| | +-+- [.--: W]
| | +--- [.-: A]
| | +--- [.---: J]
| +-+- [.-.-.-: .]
| +-+- [.--.: P]
| | +--- [.-.: R]
| +--- [.-..: L]
+-+- [...--: 3]
+-+- [..--..: ?]
| +-+- [..---: 2]
| | +--- [..-: U]
| +-+- [...: S]
| +--- [..-.: F]
| +--- [...-: V]
+-+- [....-: 4]
+--- [....: H]
+--- [.....: 5]
Pre-Order Output:
-..: D
--.-: Q
-----: 0
--: M
-: T
---: O
---..: 8
----.: 9
--.: G
-.: N
--..--: ,
--..: Z
--...: 7
-.--: Y
-.-: K
-.-.: C
..: I
.----: 1
.: E
-...: B
-..-: X
-....: 6
.--: W
.-: A
.---: J
.-.-.-: .
.--.: P
.-.: R
.-..: L
...--: 3
..--..: ?
..---: 2
..-: U
...: S
..-.: F
...-: V
....-: 4
....: H
.....: 5
In-Order Output:
-: T
--: M
---: O
-----: 0
----.: 9
---..: 8
--.: G
--.-: Q
--..: Z
--..--: ,
--...: 7
-.: N
-.-: K
-.--: Y
-.-.: C
-..: D
-..-: X
-...: B
-....: 6
.: E
.-: A
.--: W
.---: J
.----: 1
.--.: P
.-.: R
.-.-.-: .
.-..: L
..: I
..-: U
..---: 2
..--..: ?
..-.: F
...: S
...-: V
...--: 3
....: H
....-: 4
.....: 5
The Rev. Tree:
+-+- [D: -..]
+-+- [0: -----]
| +-+- [,: --..--]
| | +--- [.: .-.-.-]
| +-+- [8: ---..]
| +-+- [7: --...]
| | +-+- [1: .----]
| | +-+- [6: -....]
| | +-+- [3: ...--]
| | +--- [2: ..---]
| | +-+- [4: ....-]
| | +--- [5: .....]
| +-+- [9: ----.]
| +-+- [C: -.-.]
| +-+- [B: -...]
| +-+- [A: .-]
| +--- [?: ..--..]
+-+- [Q: --.-]
+-+- [M: --]
| +-+- [G: --.]
| | +-+- [E: .]
| | | +--- [F: ..-.]
| | +-+- [K: -.-]
| | +-+- [I: ..]
| | | +--- [H: ....]
| | | +--- [J: .---]
| | +--- [L: .-..]
| +-+- [O: ---]
| +--- [N: -.]
| +--- [P: .--.]
+-+- [T: -]
+-+- [R: .-.]
| +--- [S: ...]
+-+- [Z: --..]
+-+- [Y: -.--]
+-+- [X: -..-]
+-+- [W: .--]
+-+- [U: ..-]
+--- [V: ...-]
In-Order Output of Rev. Tree:
,: --..--
.: .-.-.-
0: -----
1: .----
2: ..---
3: ...--
4: ....-
5: .....
6: -....
7: --...
8: ---..
9: ----.
?: ..--..
A: .-
B: -...
C: -.-.
D: -..
E: .
F: ..-.
G: --.
H: ....
I: ..
J: .---
K: -.-
L: .-..
M: --
N: -.
O: ---
P: .--.
Q: --.-
R: .-.
S: ...
T: -
U: ..-
V: ...-
W: .--
X: -..-
Y: -.--
Z: --..
That's it.
$
注意：
我刚刚认识到“反转”树不再平衡。
因此，无法实现 O(ld(n)) 的最优最坏情况时间复杂度。