Question

是否可以编写一个函数来返回二叉树的给定节点的父节点？

BinaryTree *search_val(BinaryTree *bt, int val)
{
    //temp pointer 
    BinaryTree* temp = NULL;
    if(!bt->isEmpty())
    {
        //check if root is equal to value and return root if true
        if(bt->getData() == val)
        {
            return bt;
        }
        else
        {
            //search left side
            temp = search_val(bt->left(), val);
            //if not found in left, search right
            if (temp == NULL)
            {
                temp = search_val(bt->right(), val);
            }
            return temp;
        }
        return NULL;
    }
    return NULL;
 }

我现在只有这个搜索功能。我实际上是从这里得到的。所以我试图将其转换为搜索节点的父节点。参数将是根节点和我们想要的父节点。这甚至可能吗？我只需要一些提示即可开始，然后我会发布我的代码。创建此函数的目的是因为我有一个几乎完美的删除叶节点函数....唯一的问题是当我删除后打印所有节点时，仍然会出现所谓的删除节点。我确定这是因为父节点仍然在main中链接到它。这是我的删除叶节点函数：

void delete_leaf_node(BinaryTree *bt, int val)
{
    BinaryTree *temp;
    temp = search_val(bt, val);
    //If node does not exist in the tree, inform the user
    if(temp == NULL)
    {
        cout << "\n   " << val << " was not found in the tree" << endl; 
    }
    //Check if node is a leaf
    else if(temp->isLeaf())
    {
        delete temp;
        cout << "\n   Leaf " << temp->getData() << " deleted" << endl;
    }
    //Inform user that node is not a leaf
    else 
        cout << "\n   " << temp->getData() << " is not a Leaf" << endl; 
    //Display using In Order Traversal to see that the node was actually deleted    
    cout << "\n   In Order Traversal after deleting: " << endl << "\n   ";
    inOrderTraverse(bt);
    cout << endl;
}

我希望我对某人有意义......对不起，我试图缩短这个问题但不能。

BinaryTree.h文件：

using namespace std;

//BinaryTree class
class BinaryTree{
    public:
        BinaryTree();
        bool isEmpty();
        bool isLeaf();
        int getData();
        void insert(const int &DATA);
        BinaryTree *left();
        BinaryTree *right();
        void makeLeft(BinaryTree *bt);
        void makeRight(BinaryTree *bt);
    private:
        bool nullTree;
        int treeData;
        BinaryTree *leftTree;
        BinaryTree *rightTree;
};

BinaryTree.cpp文件：

#include <iostream>
#include "BinaryTree.h"

using namespace std;

//constructor
BinaryTree::BinaryTree()
{
    nullTree = true;
    leftTree = NULL;
    rightTree = NULL;
}

/*
  is_empty function for BinaryTree class. Does not take any parameters. 
  Returns true if tree is empty and false otherwise.
*/
bool BinaryTree::isEmpty()
{
    return nullTree;
}

/*
  is_leaf function for BinaryTree class. Does not take any parameters. 
  Returns true if node has no children and false otherwise.
*/
bool BinaryTree::isLeaf()
{
    return ((this->leftTree->treeData == 0) && (this->rightTree->treeData == 0));
}

/*
  getData function for BinaryTree class. Does not take any parameters. 
  Returns treeData value.
*/
int BinaryTree::getData()
{
    if(!isEmpty());
    return treeData;
}

/*
  insert function for BinaryTree class. Takes one parameter, passed by
  reference. Returns true if node has no children and false otherwise.
*/
void BinaryTree::insert(const int &DATA)
{
    //create empty children and insert DATA
    treeData = DATA;
    if(nullTree) 
    {
        nullTree = false;
        leftTree = new BinaryTree;
        rightTree = new BinaryTree;
    }
}

/*
  left function for BinaryTree class. It points to the left node.
  Does not take any parameters. Returns left node.
*/
BinaryTree *BinaryTree::left()
{
    if(!isEmpty());
    return leftTree;
}

/*
  right function for BinaryTree class. It points to the right node.
  Does not take any parameters. Returns right node.
*/
BinaryTree *BinaryTree::right()
{
    if(!isEmpty());
    return rightTree;
}

/*
  makeLeft function for BinaryTree class. Takes a pointer to a tree node as a parameter. 
  makes the parameter the left child of a node. Does not return any value
*/
void BinaryTree::makeLeft(BinaryTree *bt)
{
    if(!isEmpty());
    leftTree = bt;
}

/*
  makeRight function for BinaryTree class. Takes a pointer to a tree node as a parameter. 
  makes the parameter the right child of a node. Does not return any value
*/
void BinaryTree::makeRight(BinaryTree *bt)
{
    if (!isEmpty());
    rightTree = bt;
}

由于

Answer 1

这取决于你的BinaryTree实现。据我所知，如果你没有将每个节点内的引用保存到其父节点，则在删除时无法直接访问它

修改

您可以使用以下命令修改BinaryTree类：

class BinaryTree{ public: BinaryTree(); bool isEmpty(); bool isLeaf(); int getData(); void insert(const int &DATA); BinaryTree *left(); BinaryTree *right(); void makeLeft(BinaryTree *bt); void makeRight(BinaryTree *bt); void setParent(BinaryTree *parent); BinaryTree* getParent(); private: bool nullTree; int treeData; BinaryTree *leftTree; BinaryTree *rightTree; BinaryTree* parent; };

然后在.cpp：

BinaryTree::BinaryTree() { nullTree = true; leftTree = NULL; rightTree = NULL; parent = NULL; } void BinaryTree::setParent(BinaryTree *parent){ this->parent = parent; } BinaryTree* BinaryTree::getParent(){ return parent; }

您的删除功能如下：

void delete_leaf_node(BinaryTree *bt, int val) { BinaryTree *temp; temp = search_val(bt, val); //If node does not exist in the tree, inform the user if(temp == NULL) { cout << "\n " << val << " was not found in the tree" << endl; } //Check if node is a leaf else if(temp->isLeaf()) { // You must distinguish which child you are BinaryTree* parent = temp->getParent(); BinaryTree* leftChild = parent->left; BinaryTree* rightChild = parent->right; if(leftChild == temp){ parent->left = null; } if(rightChild == temp){ parent->right = null; } delete temp; cout << "\n Leaf " << temp->getData() << " deleted" << endl; } //Inform user that node is not a leaf else cout << "\n " << temp->getData() << " is not a Leaf" << endl; //Display using In Order Traversal to see that the node was actually deleted cout << "\n In Order Traversal after deleting: " << endl << "\n "; inOrderTraverse(bt); cout << endl; }

查找二叉树的父节点函数

1 个答案: