我试图弄清楚如何计算二叉搜索树中的叶节点数。
我不断收到运行时错误，并且 CodeBlocks 在最终返回语句处不断崩溃。我在这里看到了多个例子，但我似乎仍然无法理解我哪里出错了。 我正在尝试递归地执行此操作，但是正如我之前所说，一旦我添加了函数number_of_leaves(p -> left)+ number_of_leaves(p-> right)CodeBlocks在打印出来后停止工作：
Empty tree has 0 leaf nodes. Answer:0
Single node has 1 leaf node. Answer 1
Crashes here
.
#include <queue>
#include <stack>
#include <iostream>
#include <vector>
#include <stdlib.h>
#ifndef BINARY_SEARCH_TREE
#define BINARY_SEARCH_TREE
template<class T>
class Stack: public std::stack<T> {
public:
T pop() { T tmp = std::stack<T>::top(); std::stack<T>::pop(); return tmp; }
};
template<class T>
class Queue: public std::queue<T> {
public:
T dequeue() { T tmp = std::queue<T>::front(); std::queue<T>::pop(); return tmp; }
void enqueue(const T& el) { push(el); }
};
template<class T>
class BSTNode {
public:
BSTNode() { left = right = 0; }
BSTNode(const T& e, BSTNode<T> *l = 0, BSTNode<T> *r = 0)
{ el = e, left = l, right = r; }
T el;
BSTNode<T> *left, *right;
};
template<class T>
class BST {
public:
BST() { root = 0; }
~BST() { clear(); }
void clear() { clear(root), root = 0; }
bool is_empty() const { return root == 0; }
void preorder() { preorder(root); }
void inorder() { inorder(root); }
void postorder() { postorder(root); }
void insert(const T&);
T* search(const T& el) const { return search(root, el); }
void find_and_delete_by_copying(const T&);
void find_and_delete_by_merging(const T&);
void breadth_first();
void balance(std::vector<T>, int, int);
bool is_perfectly_balanced() const { return is_perfectly_balanced(root) >= 0; }
int number_of_leaves() const { return number_of_leaves(root); }
T* recursive_search(const T& el) const { return recursive_search(root, el); }
void recursive_insert(const T& el) { recursive_insert(root, el); }
protected:
void clear(BSTNode<T>*);
T* search(BSTNode<T>*, const T&) const;
void preorder(BSTNode<T>*);
void inorder(BSTNode<T>*);
void postorder(BSTNode<T>*);
virtual void visit(BSTNode<T>* p) // virtual allows re-definition in derived classes
{ std::cout << p->el << " "; }
void delete_by_copying(BSTNode<T>*&);
void delete_by_merging(BSTNode<T>*&);
int is_perfectly_balanced(BSTNode<T>*) const; // To be provided (A4)
int number_of_leaves(BSTNode<T>*) const; // To be provided (A4)
void recursive_insert(BSTNode<T>*&, const T&); // To be provided (P6)
T* recursive_search(BSTNode<T>*, const T&) const; // To be provided (P6)
BSTNode<T>* root;
};
#endif
template<class T>
void BST<T>::clear(BSTNode<T> *p)
{
if (p != 0) {
clear(p->left);
clear(p->right);
delete p;
}
}
template<class T>
void BST<T>::insert(const T& el)
{
BSTNode<T> *p = root, *prev = 0;
while (p != 0) { // find a place for inserting new node;
prev = p;
if (el < p->el)
p = p->left;
else
p = p->right;
}
if (root == 0) // tree is empty;
root = new BSTNode<T>(el);
else if (el < prev->el)
prev->left = new BSTNode<T>(el);
else
prev->right = new BSTNode<T>(el);
}
template<class T>
T* BST<T>::search(BSTNode<T>* p, const T& el) const
{
while (p != 0) {
if (el == p->el)
return &p->el;
else if (el < p->el)
p = p->left;
else
p = p->right;
}
return 0;
}
template<class T>
void BST<T>::inorder(BSTNode<T> *p)
{
if (p != 0) {
inorder(p->left);
visit(p);
inorder(p->right);
}
}
template<class T>
void BST<T>::preorder(BSTNode<T> *p)
{
if (p != 0) {
visit(p);
preorder(p->left);
preorder(p->right);
}
}
template<class T>
void BST<T>::postorder(BSTNode<T>* p)
{
if (p != 0) {
postorder(p->left);
postorder(p->right);
visit(p);
}
}
template<class T>
void BST<T>::delete_by_copying(BSTNode<T>*& node)
{
BSTNode<T> *previous, *tmp = node;
if (node->right == 0) // node has no right child;
node = node->left;
else if (node->left == 0) // node has no left child;
node = node->right;
else {
tmp = node->left; // node has both children;
previous = node; // 1.
while (tmp->right != 0) { // 2.
previous = tmp;
tmp = tmp->right;
}
node->el = tmp->el; // 3.
if (previous == node)
previous->left = tmp->left;
else
previous->right = tmp->left; // 4.
}
delete tmp; // 5.
}
// find_and_delete_by_copying() searches the tree to locate the node containing
// el. If the node is located, the function delete_by_copying() is called.
template<class T>
void BST<T>::find_and_delete_by_copying(const T& el)
{
BSTNode<T> *p = root, *prev = 0;
while (p != 0 && !(p->el == el)) {
prev = p;
if (el < p->el)
p = p->left;
else p = p->right;
}
if (p != 0 && p->el == el) {
if (p == root)
delete_by_copying(root);
else if (prev->left == p)
delete_by_copying(prev->left);
else
delete_by_copying(prev->right);
}
else if (root != 0)
std::cout << "el " << el << " is not in the tree" << std::endl;
else
std::cout << "the tree is empty" << std::endl;
}
template<class T>
void BST<T>::delete_by_merging(BSTNode<T>*& node)
{
BSTNode<T> *tmp = node;
if (node != 0) {
if (!node->right) // node has no right child: its left
node = node->left; // child (if any) is attached to its parent;
else if (node->left == 0) // node has no left child: its right
node = node->right; // child is attached to its parent;
else { // be ready for merging subtrees;
tmp = node->left; // 1. move left
while (tmp->right != 0) // 2. and then right as far as possible;
tmp = tmp->right;
tmp->right = // 3. establish the link between the
node->right; // the rightmost node of the left
// subtree and the right subtree;
tmp = node; // 4.
node = node->left; // 5.
}
delete tmp; // 6.
}
}
template<class T>
void BST<T>::find_and_delete_by_merging(const T& el)
{
BSTNode<T> *node = root, *prev = 0;
while (node != 0) {
if (node->el == el)
break;
prev = node;
if (el < node->el)
node = node->left;
else
node = node->right;
}
if (node != 0 && node->el == el) {
if (node == root)
delete_by_merging(root);
else if (prev->left == node)
delete_by_merging(prev->left);
else
delete_by_merging(prev->right);
}
else if (root != 0)
std::cout << "el " << el << " is not in the tree" << std::endl;
else
std::cout << "the tree is empty" << std::endl;
}
template<class T>
void BST<T>::breadth_first()
{
Queue<BSTNode<T>*> queue;
BSTNode<T> *p = root;
if (p != 0) {
queue.enqueue(p);
while (!queue.empty())
{
p = queue.dequeue();
visit(p);
if (p->left != 0)
queue.enqueue(p->left);
if (p->right != 0)
queue.enqueue(p->right);
}
}
}
template<class T>
void BST<T>::balance (std::vector<T> data, int first, int last)
{
if (first <= last) {
int middle = (first + last)/2;
insert(data[middle]);
balance(data,first,middle-1);
balance(data,middle+1,last);
}
}
template<class T>
void BST<T>::recursive_insert(BSTNode<T>*& p, const T& el)
{
if (p == 0) // Anchor case, tail recursion
p = new BSTNode<T>(el);
else if (el < p->el)
recursive_insert(p->left, el);
else
recursive_insert(p->right, el);
}
template<class T>
T* BST<T>::recursive_search(BSTNode<T>* p, const T& el) const
{
if (p != 0) {
if (el == p->el) // Anchor case, tail recursion
return &p->el;
else if (el < p->el)
return recursive_search(p->left, el);
else
return recursive_search(p->right, el);
}
else
return 0;
}
问题在这里*** 我尝试使用单独的计数器来计算节点，但它只是打印所有 0。一旦我添加 number_of_leaves(( 它就会崩溃
template<class T>
int BST<T>::number_of_leaves(BSTNode<T>*) const {
BSTNode<T> *p = root;
if(p == NULL){
return 0;
}
if(p->left == NULL && p->right==NULL){
return 1;
}
else
return number_of_leaves(p->left) + number_of_leaves(p-> right);
}
测试文件如下：
#include "BST.h"
#include <iostream>
using namespace std;
int main()
{
BST<int> a;
cout << "Empty tree has 0 leaf nodes. Answer: " << a.number_of_leaves() << endl;
a.insert(4);
cout << "Single node has 1 leaf node. Answer: " << a.number_of_leaves() << endl;
a.insert(2);
cout << "Linked list of 2 nodes has 1 leaf node. Answer: "
<< a.number_of_leaves() << endl;
a.insert(6);
cout << "Full binary tree of 3 nodes has 2 leaf nodes. Answer: "
<< a.number_of_leaves() << endl;
a.insert(3), a.insert(1), a.insert(5), a.insert(7);
cout << "Full binary tree of 7 nodes has 4 leaf nodes. Answer: "
<< a.number_of_leaves() << endl;
return 0;
}
在number_of_leaves(BSTNode<T>*)中，你丢弃传递的参数并始终从根开始。然后你递归地向下，总是做完全相同的操作，这导致了StackOverflow(对不起，我无法抗拒:p(。达到函数调用的最大数量，程序被终止。
template<class T>
int BST<T>::number_of_leaves(BSTNode<T>* start) const
{
if(start == NULL)
{
return 0;
}
if(start->left == NULL && start->right==NULL)
{
return 1;
}
return number_of_leaves(start->left) + number_of_leaves(start-> right);
}