为什么我在树的插入函数中收到分段错误？

Question

分段错误发生在insert函数中，特别是在第一个*root = x;语句中的if处。功能已经布局，我们只需要实现它们，所以我不能添加另一个参数或任何东西。我已经坚持了一段时间，并且不知道它可能是什么。根指针被分配了内存，所以我真的不知道问题是什么。感谢任何帮助。谢谢。

#include "treeNode.h"
#include <iomanip>

template <class V>
class tree {
    TreeNode<V> * root;
    TreeNode<V> * cur;
    int size;

    public:
    // default constructor                                                      
    // by default, the tree is empty                                            
    tree(){
        root = nullptr;
        size = 0;
    }

    // search value x in tree rooted at node t                                  
    bool treeSearch(V x, TreeNode<V>* t){
        if(t == nullptr) return false;
        if(t->getDatum() == x) return true;
        return treeSearch(x, t->getLeft()) || treeSearch(x, t->getRight());
    }

    bool treeSearch(V x){
        treeSearch(x, root);
    }


    // binary search value x in tree rooted at node t                           
    bool treeBSearch(V x, TreeNode<V>* t){
      if(t == nullptr) return false;
      if(t->getDatum() == x) return true;
      if(t->getDatum() > x)
        return treeBSearch(x, t->getLeft());
      else
        return treeBSearch(x, t->getRight());
    }

    bool treeBSearch(V x){
        return treeBSearch(x, root);
    }

    // check node t is leaf                                                     
    bool isLeaf(TreeNode<V>* t){
      if(t != nullptr && t->getLeft() == nullptr && t->getRight() == nullptr){
        return true;
      }else{
        return false;
      }
    }

    // find the height of the tree rooted at node t                             
 int height(TreeNode<V>* t){
      if(t == nullptr) return -1;
      if(isLeaf(t)) return 0;
      return 1 + std :: max(height(t->getLeft()), height(t->getRight()));
    }

    int height(){
        return height(root);
    }

    // find the number of nodes of tree rooted at t                             
    int nNodes(TreeNode<V>* t){
      if(t == nullptr) return 0;
      return 1 + nNodes(t->getLeft()) + nNodes(t->getRight());
    }

    int nNodes(){
        return nNodes(root);
    }


    // insert value x to the current tree object                                
    void insert(V x){
      if(root == nullptr){
        *root = x;
      }
      else{
        cur = root;
        while(cur != nullptr){
          if(cur->getDatum() < x){
            if(cur->getLeft() == nullptr){
              cur->setDatum(x);
              cur->setLeft(cur);
            }
            else{
              cur = cur->getLeft();
            }
          }
          else{
            if(cur->getRight() == nullptr){
              cur->setLeft(cur);
            }
            else{
              cur = cur->getRight();
            }
          }
        }
      }
    }

    // print out the values of tree rooted at x                                 
    // it shows the hierarchy of the tree                                       
    // it will be useful for debugging                                          
    void print(TreeNode<V> * x, int indent){
        if(x == nullptr) return;
        if (x->getRight() != nullptr) {
            print(x->getRight(), indent+4);
        }
  if (indent != 0) {
            cout << std::setw(indent) << ' ';
        }
        if(x->getRight() != nullptr){
            cout << " /\n" << std::setw(indent) << ' ';
        }
        cout << x->getDatum() << endl;
        if (x->getLeft() != nullptr) {
            cout << std::setw(indent) << ' ' <<" \\\n";
            print(x->getLeft(), indent+4);
        }
    }
    void print(){
        int count = 0;
        print(root, count);
    }
};

这是treeNode.h文件，其中包含每个函数的构造函数。

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

template <class T>
class TreeNode{
    T datum;
    TreeNode<T>* left, * right;

    public:
    // constructor with datum value, left and right are nullptr                 
    TreeNode(T x){
        datum = x;
        left = nullptr;
        right = nullptr;
    }
    // constructor with datum value, left and right values                      
    TreeNode(T x, TreeNode<T>* lft, TreeNode<T>* rgt){
        datum = x;
        left = lft;
        right = rgt;
    }

    //destructor releases left and right nodes, if not nullptr                  
    ~TreeNode(){
        if (left) {
            delete left;
        }
        if (right) {
            delete right;
        }
    }

    // get datum value                                                          
    T getDatum(){
        return datum;
    }

    // get left pointer                                                         
    TreeNode<T>* getLeft(){
        return left;
    }

    // get right pointer                                                        
    TreeNode<T>* getRight(){
        return right;
    }
    // set the left pointer                                                     
    void setLeft(TreeNode<T>* p){
        left = p;
    }
 // set the right pointer                                                    
    void setRight(TreeNode<T>* p){
        right = p;
    }
};

Answer 1

这应该是显而易见的：

 if(root == nullptr){
    *root = x;
    ^ ~~~~ dereferencing null pointer causes UB
  }

root是nullptr，但你取消它并分配 - 这是UB未定义的行为，这个特定的行为导致段错误。

如何解决它 - 你应该代替取消引用空指针，为它赋值：

  if(root == nullptr){
      root = new TreeNode<V>( x );
  }