我是这样实现的:
#include "binary_tree_with_vector.h"
#include <vector>
using namespace std;
template <typename T>
class BinaryTree {
public:
class Position {
private:
int key;
public:
Position(int k) : key(k) {}
friend class BinaryTree;
};
protected:
vector<T> *array;
public:
BinaryTree() {
array = new vector<T>;
array->push_back(T());
}
int size() {
return int(array->size()) - 1;
}
~BinaryTree() {
delete [] array;
}
BinaryTree<T>& operator=(const BinaryTree<T>& t) {
if (this != &t) {
copyFrom(t);
}
return *this;
}
BinaryTree(const BinaryTree<T>& t) {
array = new vector<T>;
copyFrom(t);
}
void print() {
cout << "size is: " << size() << endl;
for (int i = 1; i <= size(); i++) {
cout << array->at(i) << "\t";
}
cout << endl;
}
protected:
void copyFrom(const BinaryTree& t) {
for (int i = 1; i <= t.size(); i++) {
array->push_back(t[i]);
}
}
public:
void swapElements(const Position& p1, const Position& p2) {
T element = array[p1.key];
array[p2.key] = array[p1.key];
array[p1.key] = element;
}
void replaceElement(const Position& p, const T& e) {
array[p.key] = e;
}
Position root() {
return Position(array[1]);
}
bool isRoot(const Position& p) {
return p.key ==1;
}
bool isLeft(const Position& p) {
return p.key % 2 == 0;
}
Position parent(const Position& p) {
return Position(array->at(p.key / 2));
}
Position leftChild(const Position& p) {
return Position(array->at(p.key * 2));
}
Position rightChild(const Position& p) {
return Position(array->at(p.key * 2 + 1));
}
Position sibling(const Position& p) {
if (isLeft(p)) {
return Position(array->at(p.key + 1));
}
return Position(array->at(p.key - 1));
}
bool isExternal(const Position& p) {
return p.key * 2 > size();
}
bool isInternal(const Position& p) {
return !isExternal(p);
}
Position insert(const T& e) {
array->push_back(e);
return
(size());
}
T elementOf(const Position& p) {
return array->at(p.key);
}
};
void binary_tree_with_vector() {
typedef BinaryTree<int>::Position Position;
BinaryTree<int> tree;
Position root = tree.insert(1);
tree.insert(2);
tree.insert(3);
tree.insert(4);
tree.insert(5);
tree.insert(6);
tree.insert(7);
tree.print();
//1
//2 3
//4 5 6 7
Position leftChild = tree.leftChild(root);
cout << "left child of root is: " << tree.elementOf(leftChild) << endl;
Position rightChild = tree.rightChild(leftChild);
cout << "right child of left child of root is: " << tree.elementOf(rightChild) << endl;
Position rightLeft = tree.leftChild(tree.rightChild(root));
cout << "right left is: " << tree.elementOf(rightLeft) << endl;
}
位置只是围绕等级(键或索引)的封装,因为树的实现细节应该隐藏在外面。
当我运行它时,我在析构函数中得到了错误
size is: 5
chap6(607,0x7fff7bea6310) malloc: *** error for object 0x100103a88: pointer being freed was not allocated
*** set a breakpoint in malloc_error_break to debug
1 2 3 4 5 (lldb)
最佳答案
你在这里分配了一个vector
array = new vector<T>;
但正在尝试删除
此处的vector
数组
delete [] array;
将最后一行改为
delete array;
关于c++数据结构用 vector 实现二叉树,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23287010/