二叉查找树 - 根据二叉树的特点进行查找

2017-07-22  20:35:51

writer：pprp

在创建二叉树的基础上进行查找，由于二叉树的特点最快为O(logn)，最慢为O(n)。

代码如下：

#include <iostream>

using namespace std;

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



void print(tree * root)
{
    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)
{
    tree *root = NULL;
    for(int i = 0 ; i< len ; i++ )
    {
        root = insert(root,node[i]);
    }
    return root;
}

tree* insert(tree * root,int key)
{
    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)
{
//    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;
    }
  
    return root;

}

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 << "您想查找的节点大小：" << endl;
    cin >> x;

    point = bisearch(root,x);

    if(point == NULL)
        cout <<"没有您要的数" << endl;
    else
        cout <<point->date<< endl;
    return 0;
}