Tree Traversals Again
1086.Tree Traversals Again (25)
An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example,suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed,the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.
Input Specification:
Each input file contains one test case. For each case,the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow,each describes a stack operation in the format: “Push X” where X is the index of the node being pushed onto the stack; or “Pop” meaning to pop one node from the stack.
Output Specification:
For each test case,print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space,and there must be no extra space at the end of the line.
Sample Input:
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
Sample Output:
3 4 2 6 5 1
题目大意：
用栈的形式给出一棵二叉树的建立的顺序，求这棵二叉树的后序遍历
分析：栈实现的是二叉树的中序遍历（左根右），而每次push入值的顺序是二叉树的前序遍历（根左右），所以该题可以用二叉树前序和中序转后序的方法做~~
#include <cstdio>
#include <stack>
using namespace std;
int preorder[35],inorder[35];
int n,preid = 0,inid = 0,cnt = 0;
int get(){
char s[10];
scanf("%s",s);
if (s[1] == 'o') return -1;
int a;
scanf("%d",&a);
return a;
}
void build(int preb,int pree,int inb,int ine){
if (preb > pree) return;
int root = preorder[preb];
int inroot = inb;
while (inorder[inroot] != root) ++inroot;
build(preb+1,preb+inroot-inb,inb,inroot-1);
build(preb+inroot-inb+1,pree,inroot+1,ine);
if (cnt++ != 0) putchar(' ');
printf("%d",root);
}
int main(){
scanf("%d",&n);
stack<int> st;
for (int i = 0; i < n*2; ++i){
int a = get();
if (a != -1){
st.push(a);
preorder[preid++] = a;
}else{
inorder[inid++] = st.top();
st.pop();
}
}
build(0,n-1,0,n-1);
return 0;
}
Complete Binary Search Tree
A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:
The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the left and right subtrees must also be binary search trees.
A Complete Binary Tree (CBT) is a tree that is completely filled,with the possible exception of the bottom level,which is filled from left to right.
Now given a sequence of distinct non-negative integer keys,a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.
Input Specification:
Each input file contains one test case. For each case,the first line contains a positive integer N (<=1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.
Output Specification:
For each test case,print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space,and there must be no extra space at the end of the line.
Sample Input:
10
1 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4
题意：
给一串构成树的序列，已知该树是完全二叉搜索树，求它的层序遍历的序列
分析：
总得概括来说，已知中序，可求root下标，可以求出层序。
1. 因为二叉搜索树的中序满足：是一组序列的从小到大排列，所以只需排序所给序列即可得到中序
2. 因为根据完全二叉树的结点数，可以求出它的根结点在中序中对应的下标
3. 如此，已知了中序，又可以根据结点数求出根结点的下标，就可以递归求出左右子树的根结点的下标
4. i结点的左孩子为2 * i + 1，右孩子2 * i + 2，就可以根据结点下标和中序数组赋值level数组
5. 最后输出所有结点的层序数组level
图1
图2
图3
#include <cstdio>
#include <cstdlib>
#include <math.h>
const int MaxNum = 1005;
int node[MaxNum];
int tree[MaxNum];
int cmp(const void *a,const void *b)
{/*以图2为基准*/
int *pa = (int *)a;
int *pb = (int *)b;
return *pa-*pb;
}
int GetLeftLength(int n)
{/*以图3为基准*/
int h,x,l;
/*转换成以2为底*/
h = log((double)(n + 1))/log (2.0);
x = n + 1 - pow(2.0,h);
if (x > pow(2.0,h - 1)) {
x = pow(2.0,h - 1);
}
l = pow(2.0,h - 1) - 1 + x;
return l;
}
void solve(int ALeft,int ARight,int TRoot)
{/*初始调用为solve(0,N-1,0)*/
/*以图1为基准*/
int n,L,LeftTRoot,RightTRoot;
n = ARight - ALeft + 1;
if(n==0) return; /*考虑完美二叉树的情况*/
L = GetLeftLength(n); /*计算出n个节点的树，其左子树有多少个节点*/
tree[TRoot] = node[ALeft + L];
LeftTRoot = TRoot * 2 + 1;
RightTRoot = LeftTRoot + 1;
solve(ALeft,ALeft + L -1,LeftTRoot);
solve(ALeft + L + 1,ARight,RightTRoot);
}
int main()
{
int i,n;
/*输入*/
scanf("%d",&n);
for(i=0;i<n;++i){
scanf("%d",&node[i]);
}
/*从大到小排序*/
qsort(node,n,sizeof(int),cmp);
/*排序成完全二叉树*/
solve(0,0);
/*输出*/
printf("%d",tree[0]);
for(i=1;i<n;++i){
printf(" %d",tree[i]);
}
printf("\n");
system("pause");
return 0;
}
Huffman Codes
PTA - Data Structures and Algorithms (English) - 5-9
In 1953,David A. Huffman published his paper “A Method for the Construction of Minimum-Redundancy Codes”,and hence printed his name in the history of computer science. As a professor who gives the final exam problem on Huffman codes,I am encountering a big problem: the Huffman codes are NOT unique. For example,given a string
Input Specification:
Each input file contains one test case. For each case,the first line gives an integer N
where
where
Output Specification:
For each test case,print in each line either “
Note: The optimal solution is
Sample Input:
7 //结点数目num
A 1 B 1 C 1 D 3 E 3 F 6 G 6 //每个结点数据data及出现的次数weight
4 //测试数据的组数checkNum
A 00000 //之后的 4*7行 是结点数据ch及其编码s
B 00001
C 0001
D 001
E 01
F 10
G 11
A 01010
B 01011
C 0100
D 011
E 10
F 11
G 00
A 000
B 001
C 010
D 011
E 100
F 101
G 110
A 00000
B 00001
C 0001
D 001
E 00
F 10
G 11
Sample Output:
Yes
Yes
No
No
题目分析：
这是一道考察“哈夫曼编码”的问题，但是这里不一定非要把哈夫曼树构造出来。Note: The optimal solution is not necessarily generated by Huffman algorithm
- 输入：第一行是结点数目num；第二行是每个结点数据data及出现的次数weight；第三行是测试数据的组数checkNum；第四行及以后是结点数据ch及编码s。
例1：最优编码不一定通过Huffman算法得到。给定4个字符及其出现频率：
A. A:000; B:001; C:01; D:1
B. A:10; B:11; C:00; D:01
C. A:00; B:10; C:01; D:11
D. A:111; B:001; C:10; D:1
解析：C
根据答案，画出每个答案的Huffman tree，再一个个的对答案是否正确。
例2：Huffman Codes的特点：
- 最优编码 —— 总长度（WPL）最小；
- 无歧义解码 —— 前缀码：数据仅存于叶子结点；
- 没有度为1 的结点 ——
满足1、2则必然有3 。
注意：满@H_502_2763@足2、3可不一定有1 ！比如：
解法：
解法转自：http://www.cnblogs.com/clevercong/p/4193370.html
1) map 用于存放：A 1 B 1 C 1 D 3 E 3 F 6 G 6 //每个结点的数据data及出现的次数(权值)weight
2) 使用C++标准库中的优先队列：priority_queue，引入头文件 #include 。优先队列底层由堆实现，数据放入队列后，会自动按照“优先级”排好顺序。
#include <map>
#include <queue>
map<char,int> myMap;
priority_queue<int,vector<int>,greater<int> >pq; //小顶堆
for(int i=0; i<num; i++) // 输入结点的数据c[i]、权值f[i]
{
cin >> c[i] >> f[i];
myMap[c[i]] = f[i]; // 映射
pq.push(f[i]); // 向队列中添加元素
}
3） 计算
// 计算WPL的值
int myWpl = 0;
while(!pq.empty())
{
int myTop = pq.top();
pq.pop();
if(!pq.empty())
{
int myTop2 = pq.top();
pq.pop();
pq.push(myTop + myTop2);
int m = myTop + myTop2;
myWpl += m; //每次加m(子节点权值重复加入) 等效于 路径长度*权值
}
}
4) 测试数据需按编码排序，但标准库并没有为map制定sort函数，因此我们用vector装载pair类型，既可以模仿出map的功能，又可以用vector的排序函数。
#include <algorithm> // sort()
typedef pair<char,string> PAIR; // 用PAIR来代替pair<char,string> (编码类型:string)
// cmp()：自定义按什么内容或大小顺序排序
// 这里是按编码的长度排序
int cmp(const PAIR& x,const PAIR& y)
{
return x.second.size() < y.second.size();
}
// vector + pair<,> 模仿 map
vector<PAIR> checkVec;
checkVec.push_back(make_pair(ch,s)); // 向vector中添加元素
sort(checkVec.begin(),checkVec.end(),cmp); // 按照编码的长度排序
5) 判断前缀问题：substr函数，取字符串中的一段并与当前编码进行比较。
bool flag = true; //已符合条件一：wpl最小
for(int i=0; i<num; i++)
{
string tmp = checkVec[i].second;
for(int j=i+1; j<num; j++)
{
if(checkVec[j].second.substr(0,tmp.size())==tmp)
flag = false;
}
}
完整代码：
#include <iostream>
#include <algorithm> // 排序函数 sort()
#include <map>
#include <queue>
using namespace std;
typedef pair<char,string> PAIR; // + vector来模仿 map
int cmp(const PAIR& x,const PAIR& y) // 自定义让sort()按哪种方式排序
{
return x.second.size() < y.second.size();
}
int main()
{
int num;
cin >> num;
char *c = new char[num];
int *f = new int[num];
map<char,int> myMap; // 用来存节点数据及权值，并构成映射
// 使用优级队列
priority_queue<int,greater<int> >pq;
for(int i=0; i<num; i++) // 输入结点及出现次数(权值)
{
cin >> c[i] >> f[i];
myMap[c[i]] = f[i];
pq.push(f[i]); // 将权值压入优先队列
}
// 计算WPL的值
int myWpl = 0;
while(!pq.empty())
{
int myTop = pq.top();
pq.pop();
if(!pq.empty())
{
int myTop2 = pq.top();
pq.pop();
pq.push(myTop + myTop2);
int m = myTop + myTop2;
myWpl += m;
}
}
// 输入测试数据
int checkNum; // 测试数据的组数
cin >> checkNum;
for(int i=0; i<checkNum; i++)
{
int wpl = 0;
char ch;
string s;
// vector + PAIR 模仿 map，使其可排序
vector<PAIR> checkVec;
for(int j=0; j<num; j++)
{
cin >> ch >> s;
checkVec.push_back(make_pair(ch,s)); // 向vector中添加测试数据及其编码
wpl += s.size() * myMap[ch];
}
sort(checkVec.begin(),cmp); // 按照编码长度排序
if(wpl != myWpl)
{
cout << "No" << endl;
continue;
}
else
{
bool flag = true; // 表示已满足条件一：wpl最小(wpl==myWpl)
//条件二：编码的前缀不能是其他编码的前缀：substr()
for(int i=0; i<num; i++)
{
string tmp = checkVec[i].second;
for(int j=i+1; j<num; j++)
{
if(checkVec[j].second.substr(0,tmp.size())==tmp)
flag = false;
}
}
if(flag == true)
cout << "Yes" << endl;
else
cout << "No" << endl;
continue;
}
cout << "Yes" << endl;
}
return 0;
}
总结的程序
#include <iostream>
#include <string>
using namespace std;
struct Heap
{
int *data;
int size = 0;
};
struct cnode
{
int tag = 0;
cnode *left = nullptr;
cnode *right = nullptr;
};
Heap *Init_Heap(int n); //初始化小根堆（利用静态链表的逻辑结构）；
void Insert_Heap(Heap *H,int x); //把元素依次插入小根堆；
int Delmin_Heap(Heap *H); //删除小根堆中的最小元素；
int Cal_Wpl(Heap *H); //计算huffman编码的wpl；
int wpl_prefix(int *sample,int n,int &re); //计算待测编码的wpl及判断是否为前缀编码；
/*思路：要判断是否，需要解决两个问题： 1）编码wpl等于huffman编码的wpl； 2)待测编码是前缀编码。 问题1： 首先要求出标准wpl。观察huffman树，我们发现其wpl是非叶子结点权值和。 于是，我们无需构造出huffman树来求权值(麻烦点），通过模拟树的构造过程， 理由小根堆的特点求出wpl； 而待测编码的wpl就是利用每个编码的字符串长度，乘以每个字符的权值得到； 问题2： 思路是构造一个二叉树出来，模拟huffman树。 1.让每个编码遍历树，结束时在当前结点设置标记为1； 2.如果遍历时，未结束时碰到了标记为1的结点，则不是前缀编码； 3.结束时，未碰到标记点（包括最后一个），是前缀编码。 */
int main()
{
int n,i,val;
char ch;
cin >> n;
int *sample = new int[n]; //每个字符权值存储在这里；
Heap *H = Init_Heap(n); //初始化小根堆（利用静态链表的逻辑结构）；
for (i = 0; i < n; i++)
{
cin >> ch >> val;
sample[i] = val;
Insert_Heap(H,val); //把元素依次插入小根堆；
}
int best_wpl = Cal_Wpl(H); //计算huffman编码的wpl；
int m,wpl,re; //re是用来判断是否前缀编码，是为1，否为0；
cin >> m;
for (i = 0; i < m; i++)
{
wpl = wpl_prefix(sample,re); //计算待测编码的wpl及判断是否为前缀编码；
if (wpl == best_wpl && re == 1)
cout << "Yes" << endl;
else
cout << "No" << endl;
}
delete sample;
return 0;
}
Heap *Init_Heap(int n)
{
Heap *H = new Heap;
H->data = new int[n + 1]; //堆的下标从1开始（为了操作方便）；
H->data[0] = -1; //小根堆赋值最小值；
return H;
}
void Insert_Heap(Heap *H,int x) //把元素依次插入小根堆；
{
int i = ++(H->size);
for (; H->data[i / 2] > x; i /= 2) //从下往上过滤；
H->data[i] = H->data[i / 2];
H->data[i] = x;
return;
}
int Delmin_Heap(Heap *H) //删除小根堆中的最小元素；
{
int t = H->data[H->size--];
int min = H->data[1];
int pa,chi;
for (pa = 1; pa * 2 <= H->size; pa = chi) //从上往下过滤最小元素；
{
chi = pa * 2;
if (chi != H->size && H->data[chi + 1] < H->data[chi]) //找到子结点的最小下标；
chi++;
if (t < H->data[chi])
break;
else
H->data[pa] = H->data[chi];
}
H->data[pa] = t; //赋值最小元素
return min;
}
int Cal_Wpl(Heap *H) //计算huffman编码的wpl；
{
int sum = 0;
if (H->size > 1)
{
int i,t1,t2,t;
int m = H->size;
for (i = 1; i < m; i++)
{
t1 = Delmin_Heap(H); //模拟构造huffman树的过程，先取出两个最小值，相加，把和插入堆中；
t2 = Delmin_Heap(H);
t = t1 + t2;
Insert_Heap(H,t);
sum += t;
}
}
else
sum = H->data[0];
return sum;
}
int wpl_prefix(int *sample,int &re) //计算待测编码的wpl及判断是否为前缀编码；
{
int i,j,len,wpl = 0;
char ch;
string st;
cnode *phead = new cnode;
cnode *p = phead;
re = 1;
for (i = 0; i < n; i++)
{
cin >> ch >> st; //此处计算wpl；
len = st.length();
wpl += len*sample[i];
if (re != 0) //此处判断是否前缀编码；
{
for (j = 0; j < len; j++)
{
if (st[j] == '0') //0的话判断左边；
{
if (p->left == nullptr) //左边空，新建结点；
{
cnode *q = new cnode;
p->left = q;
}
else
{
if (p->tag == 1)
{
re = 0;
break;
}
}
p = p->left;
}
else if (st[j] == '1') //判断右边；
{
if (p->right == nullptr)
{
cnode *q = new cnode;
p->right = q;
}
else
{
if (p->tag == 1)
{
re = 0;
break;
}
}
p = p->right;
}
}
if (p->left == nullptr && p->right == nullptr)
re = 1;
else
re = 0;
if (p->tag == 0) //注意此处需要判断，最后结点标记不为1才能赋值（待测编码有可能有相同的）；
p->tag = 1;
else
re = 0;
p = phead;
}
}
return wpl;
} 原文链接：https://www.f2er.com/datastructure/382253.html