用C ++程序搜索二进制搜索树中的元素
在此程序中，我们需要。实现二进制搜索以在二进制搜索树中找到搜索序列的存在。二元搜索的最坏情况时间复杂度为O(n)，但对于平均情况为O（log(n)）。
算法
Begin Construct binary search tree for the given unsorted data array by inserting data into tree one by one. Take the input of data to be searched in the BST. Now starting from the root node, compare the data with data part of the node. if data < temp->d, move the temp pointer to the left child. if data > temp->d move the temp pointer to the right child. if data = temp->d print the tree depth where it is found and return to main. Else print item not found. End
范例程式码
#include<iostream>
using namespace std;
struct node {
int d;
node *left;
node *right;
};
node* CreateNode(int d) {
node *newnode = new node;
newnode->d = d;
newnode->left = NULL;
newnode->right = NULL;
return newnode;
}
node* InsertIntoTree(node* root, int d) {
node *temp = CreateNode(d);
node *t = new node;
t = root;
if(root == NULL)
root = temp;
else {
while(t != NULL) {
if(t->d < d) {
if(t->right == NULL) {
t->right = temp;
break;
}
t = t->right;
} else if(t->d > d) {
if(t->left == NULL) {
t->left = temp;
break;
}
t = t->left;
}
}
}
return root;
}
void Search(node *root, int d) {
int depth = 0;
node *temp = new node;
temp = root;
while(temp != NULL) {
depth++;
if(temp->d == d) {
cout<<"\nitem found at depth: "<<depth;
return;
} else if(temp->d > d)
temp = temp->left;
else
temp = temp->right;
}
cout<<"\n item not found";
return;
}
int main() {
char ch;
int n, i, a[10] = {93, 53, 45, 2, 7, 67, 32, 26, 71, 76};
node *root = new node;
root = NULL;
for (i = 0; i < 10; i++)
root = InsertIntoTree(root, a[i]);
up:
cout<<"\nEnter the Element to be searched: ";
cin>>n;
Search(root, n);
cout<<"\n\n\tDo you want to search more...enter choice(y/n)?";
cin>>ch;
if(c == 'y' || c == 'Y')
goto up;
return 0;
}输出结果
Enter the Element to be searched: 26 item found at depth: 7 Do you want to search more...enter choice(y/n)? Enter the Element to be searched: 1 item not found Do you want to search more...enter choice(y/n)?