3374 数据结构实验之查找二：平衡二叉树

3374 数据结构实验之查找二：平衡二叉树

Time Limit: 400MS Memory Limit: 65536KB

Problem Description

根据给定的输入序列建立一棵平衡二叉树，求出建立的平衡二叉树的树根。

Input

输入一组测试数据。数据的第1行给出一个正整数N(n <= 20)，N表示输入序列的元素个数；第2行给出N个正整数，按数据给定顺序建立平衡二叉树。

Output

输出平衡二叉树的树根。

Example Input

588 70 61 96 120

Example Output

70

#include <bits/stdc++.h>using namespace std;typedef struct node{    int data,length;    struct node *l,*r;}*tree;int n,m,flag;int length(tree root){    if(root==NULL)        return -1;    else        return root->length;}tree RR(tree root){    struct node *temp = root->r;    root->r = temp->l;    temp->l = root;    temp->length = max(length(temp->l),length(temp->r))+1;    root->length = max(length(root->l),length(root->r))+1;    return temp;}tree LL(tree root){    node *temp = root->l;    root->l = temp->r;    temp->r = root;    temp->length = max(length(temp->l),length(temp->r))+1;    root->length = max(length(root->l),length(root->r))+1;    return temp;}tree RL(tree root){    root->r = LL(root->r);    return RR(root);}tree LR(tree root){    root->l = RR(root->l);    return LL(root);}tree creat(int x,tree root){    if(root==NULL)    {        root = new node;        root->data = x;        root->length =0;        root->l = root->r = NULL;    }    else if(x>root->data)    {        root->r = creat(x,root->r);        if(length(root->r)-length(root->l)>1)        {            if(root->r->data<x)                root = RR(root);            else                root = RL(root);        }    }    else if(x<root->data)    {        root->l = creat(x,root->l);        if(length(root->l)-length(root->r)>1)        {            if(root->l->data<x)                root = LR(root);            else                root = LL(root);        }    }    root->length = max(length(root->l),length(root->r))+1;    return root;}int main(){    int x;    cin>>n;    struct node *root = NULL;    for(int i =0; i<n; i++)    {        cin >> x;        root = creat(x,root);    }    cout << root->data <<endl;;    return 0;}