二叉查找树

2017-07-23 09:09:19

writer：pprp

二叉查找树，删除的功能，分为三种情况，

1. 没有子节点
2. 有一个子节点
3. 有两个子节点 （有两种方法，具体见代码）

代码如下：

#include <iostream>

using namespace std;

struct tree
{
    tree*left;
    int date;
    tree*right;
};



void print(tree * root)   //用中序打印出来
{
    if(root!=NULL)
    {
        print(root->left);
        cout << root->date << endl;
        print(root->right);
    }
}
tree* insert(tree * root,int key);
tree * create(int *node,int len)    //建立一个二叉树
{
    tree *root = NULL;
    for(int i = 0 ; i< len ; i++ )
    {
        root = insert(root,node[i]);
    }
    return root;
}

tree* insert(tree * root,int key)   //在二叉树中根据规则插入一个数
{
    tree*current;
    tree*parent;
    tree*newval=new tree();

    newval->left = NULL;
    newval->right = NULL;
    newval->date = key;

    if(root == NULL)
    {
        root = newval;
    }
    else
    {
        current = root;
        while(current!=NULL)
        {
            parent = current;
            if(current->date>key)
            {
                current  = current->left;
            }
            else
            {
                current = current->right;
            }
        }
        if(parent->date > key)
        {
            parent->left = newval;
        }
        else
        {
            parent->right = newval;
        }
    }
    return root;
}

tree * bisearch(tree* root,int x)    //在二叉树中查找一个数
{
//    if(root!=NULL)
//    {
//        if(root->date == x)
//        {
//            return root;
//        }
//        else if(root->date > x)
//        {
//            root = bisearch(root->left,x);
//        }
//        else
//        {
//            root = bisearch(root->right,x);
//        }
//    }
//    return NULL;
    bool solve = false;
    while(root && !solve)
    {
        if(x == root->date)
            solve = true;
        else if(x < root->date)
            root = root->left;
        else
            root = root->right;
    }
    if(root == NULL)
        cout <<"can't find it" << endl;
    return root;

}

bool Delete(tree*root,int x)     //从二叉树中删除一个数
{
    bool find = false;

    tree*parent = NULL;
    tree*current = NULL;

    current = root;
    while(current&&!find)     //用parent记录下来现在节点的父节点，通过这个while循环找到x
    {
        if(current->date < x)
        {
            parent = current;
            current = current->right;
        }

        else if(current->date > x)
        {
            parent = current;
            current = current->left;
        }
        else if(current->date == x)
        {
            find = true;
        }
    }
    if(current == NULL)
        cout << "can't find " << x << endl;
    if(current->left == NULL && current->right == NULL)  //第一种情况，如果没有子节点
    {
        if( current == root)
            root = NULL;
        if(parent->left == current)
            parent->left = NULL;
        else
            parent->right = NULL;
        delete current;
    }
    else if(current->right == NULL || current->left == NULL)   //第二种情况，如果只有一个子节点
    {
        if(current == root)
        {
            if(current->left == NULL)
                root = current->right;
            else
                root = current->left;
        }
        else                                       //分四种情况，手动画图看看
        {
            if(parent->left == current && current->left)
                parent->left = current->left;
            else if(parent->left == current && current->right)
                parent->left = current->right;
            else if(parent->right == current && current->right)
                parent->right = current->right;
            else
                parent->right = current->left;
        }
        delete current;
    }
    else  //有两个子节点，可以有两种方法，
        //1，找到前继忠最大的点，交换；
        //2，找到后集中最小的点，交换；
        //这里采用的是第一种方案
    {
        tree * par = current;
        tree * kid = current->left;
        while(kid->right)
        {
            par = kid;
            kid = kid->right;
        }
        current->date = kid->date;
        if(par == current)
            current->left = kid->left;
        else
            par->right = kid->left;
        delete kid;
    }
    return find;
}

int main()
{
    int x;
    tree * root = NULL;
    tree * point = NULL;
    int node[11] = {1,2,3,4,5,6,7,8,9,10,11};
    root = create(node,11);


    print(root);


    cout << "您想查找的节点的值：" << endl;
    cin >> x;

    point = bisearch(root,x);

    if(point == NULL)
        cout <<"没有您要的数" << endl;
    else
        cout <<point->date<< endl;

    cout << "您想删除的数： " << endl;

    cin >> x;
    if(Delete(root,x) == true)
        cout <<"the number "<<x << " has been delete successfully!" << endl;
    else
        cout <<"can't find the number " <<x << endl;

    print(root);
    return 0;
}