来源于沈钰S同学(舒姐)
来源于沈钰S同学(舒姐)
abyssfall的博客_优快云博客-计科学子的头发账本,代码存档领域博主
问题描述 :
目的:现有两个结点序列,分别是对同一个二叉树进行前序遍历和中序遍历(或中序和后序)的结果。要求设计一个算法,重构该二叉树,并输出该二叉树按照后序(前序)遍历时的结点序列。假定二叉树所有的结点的数据域的值都不相同。二叉树的存储结构的建立参见二叉树应用1。
一、已知前序和中序的结果,进行二叉树的重构:
提示:preOrder按照先根再左再右的顺序递归遍历,inOrder按照先左再根再右的顺序递归遍历。
举例说明:preOrder的输入pre={A,B,D,G,H,C,E,I,F},inOrder的输入in={G,D,H,B,A,E,I,C,F}。首先按preOrder遍历的顺序依次访问各结点。访问过程中,通过in得知各子树内inOrder遍历的顺序,从而重建以当前访问结点c为根的左子树与右子树。即:设preOrder遍历的当前结点为c,c在in中的位置为m,m左侧就是c的左子树,右侧就是c的右子树。同理递归。
(2)
参考函数原型:
(1)//重建二叉树的存储结构 (外壳)
template<class ElemType>
BinaryTreeNode<ElemType> *ReBuildTree_Pre_In(vector[ElemType] &pre, vector[ElemType] &in);
(2)//重建二叉树的存储结构 (递归)
template<class ElemType>
BinaryTreeNode<ElemType> *reConstructCore_Pre_In(vector[ElemType] &pre, vector[ElemType] &in, int preStart, int preEnd, int inStart, int inEnd );
二、已知中序和后序的结果,进行二叉树的重构:
可参考一的思路。
参考函数原型:
(1)//重建二叉树的存储结构 (外壳)
template<class ElemType>
BinaryTreeNode<ElemType> *ReBuildTree_In_Post(vector[ElemType] &in, vector[ElemType] &post);
(2)//重建二叉树的存储结构 (递归)
template<class ElemType>
BinaryTreeNode<ElemType> *reConstructCore_In_Post(vector[ElemType] &in, vector[ElemType] &post, int inStart, int inEnd, int postStart, int postEnd );
输入说明 :
第一行:重构选择(0:前序中序重构 1:中序后序重构)
0:
第二行:按前序遍历的结点序列,相邻结点用空格隔开
第三行:按中序遍历的结点序列,相邻结点用空格隔开
1:
第二行:按中序遍历的结点序列,相邻结点用空格隔开
第三行:按后序遍历的结点序列,相邻结点用空格隔开
输出说明 :
0:
第一行:按后序遍历的结点序列,相邻结点用","隔开
1:
第一行:按前序遍历的结点序列,相邻结点用","隔开
-----------
0
A B C D E
B D C E A
----------------
D,E,C,B,A
#include <bits/stdc++.h>
using namespace std;
template <class ElemType>
struct BinaryTreeNode
{
ElemType data;
BinaryTreeNode<ElemType> *left;
BinaryTreeNode<ElemType> *right;
};
template <class ElemType>
void cinin(string str, vector<ElemType> &a)
{
int len = str.length();
string str0 = "";
for (int i = 0; i < len; i++)
{
if (str[i] != ' ')
{
str0 += str[i];
}
else
{
a.push_back(str0);
str0 = "";
}
}
a.push_back(str0);
}
bool f = false;
//前序遍历
template <class ElemType>
void visit1(BinaryTreeNode<ElemType> *head)
{
if (head)
{
if (f == false)
{
f = true;
cout << head->data;
}
else
{
cout << "," << head->data;
}
}
if (head->left)
visit1(head->left);
if (head->right)
visit1(head->right);
return;
}
//后序遍历
template <class ElemType>
void visit3(BinaryTreeNode<ElemType> *head)
{
if (head->left)
visit3(head->left);
if (head->right)
visit3(head->right);
if (head)
{
if (f == false)
{
f = true;
cout << head->data;
}
else
{
cout << "," << head->data;
}
}
return;
}
template <class ElemType>
BinaryTreeNode<ElemType> *reConstructCore_Pre_In(vector<ElemType> &pre, vector<ElemType> &in)
{
if (pre.size() == 0)
{
return NULL;
} //先序读完了,不进下一层递归了,没叶子了
vector<ElemType> left_pre, right_pre, left_in, right_in; //总之分一下,左右子树咔咔一下准备进递归
BinaryTreeNode<ElemType> *head = new BinaryTreeNode<ElemType>;
head->data = pre[0];
int root = 0;
int inlen = in.size();
for (int i = 0; i < inlen; i++)
{
if (pre[0] == in[i])
{
root = i;
break;
}
else
continue;
} //找一下根结点
for (int i = 0; i < root; i++)
{
left_in.push_back(in[i]); //中序遍历,根结点之前全是左子树
left_pre.push_back(pre[i + 1]); //先序遍历,第一个是根节点,往后读取相同数量丢入左子树
}
for (int i = root + 1; i < inlen; i++)
{
right_in.push_back(in[i]); //中序遍历,根节点之后全是右子树
right_pre.push_back(pre[i]); //先序遍历,往后读取相同数量丢入左子树
}
head->left = reConstructCore_Pre_In(left_pre, left_in); //新分好的左子树
head->right = reConstructCore_Pre_In(right_pre, right_in); //新分好的右子树
return head;
}
template <class ElemType>
BinaryTreeNode<ElemType> *reConstructCore_In_Post(vector<ElemType> &in, vector<ElemType> &post)
{
if (post.size() == 0)
{
return NULL;
} //先序读完了,不进下一层递归了,没叶子了
vector<ElemType> left_in, right_in, left_post, right_post; //总之分一下,左右子树咔咔一下准备进递归
BinaryTreeNode<ElemType> *head = new BinaryTreeNode<ElemType>;
int postlen = post.size();
head->data = post[postlen - 1]; //后序遍历末尾为根节点
int root = 0;
int inlen = in.size();
for (int i = 0; i < inlen; i++)
{
if (post[postlen - 1] == in[i])
{
root = i;
break;
}
else
continue;
} //找一下根结点在中序遍历的位置,分左右子树
//cout << "root = " << root << "in[root] = " << in[root] << endl;
for (int i = 0; i < root; i++)
{
left_in.push_back(in[i]); //中序遍历,根结点之前全是左子树
left_post.push_back(post[i]); //后序遍历,往后读取相同数量丢入左子树
}
//for (int i = 0; i < root; i++)
//{
// cout << "left_in:" << endl;
// cout << left_in[i] << " ";
//}
///cout << endl;
//for (int i = 0; i < root; i++)
//{
// cout << "left_post:" << endl;
// cout << left_post[i] << " ";
//}
//cout << endl;
for (int i = root + 1; i < inlen; i++)
{
right_in.push_back(in[i]); //中序遍历,根节点之后全是右子树
right_post.push_back(post[i - 1]); //后序遍历,挨着上个循环结尾往后读取相同数量丢入左子树
}
//for (int i = 0; i < right_in.size(); i++)
//{
// cout << "right_in:" << endl;
// cout << right_in[i] << " ";
//}
//cout << endl;
//for (int i = 0; i < right_post.size(); i++)
//{
// cout << "right_post:" << endl;
// cout << right_post[i] << " ";
//}
//cout << endl;
head->left = reConstructCore_In_Post(left_in, left_post);
head->right = reConstructCore_In_Post(right_in, right_post);
return head;
}
int main()
{
char flag;
cin >> flag;
getchar();
if (flag == '0')
{
string linepre;
vector<string> pre;
getline(cin, linepre);
cinin(linepre, pre);
vector<string> in;
string linein;
getline(cin, linein);
cinin(linein, in);
BinaryTreeNode<string> *root = new BinaryTreeNode<string>;
root = reConstructCore_Pre_In(pre, in);
visit3(root);
}
else if (flag == '1')
{
string linein;
vector<string> in;
getline(cin, linein);
cinin(linein, in);
string linepost;
vector<string> post;
getline(cin, linepost);
cinin(linepost, post);
BinaryTreeNode<string> *root = new BinaryTreeNode<string>;
root = reConstructCore_In_Post(in, post);
visit1(root);
}
getchar();
getchar();
return 0;
}