数据结构上机测试4.1:二叉树的遍历与应用1

Time Limit: 1000 ms Memory Limit: 65536 KiB
Submit Statistic Discuss
Problem Description
输入二叉树的先序遍历序列和中序遍历序列，输出该二叉树的后序遍历序列。
Input
第一行输入二叉树的先序遍历序列；
第二行输入二叉树的中序遍历序列。
Output
输出该二叉树的后序遍历序列。
Sample Input
ABDCEF
BDAECF
Sample Output
DBEFCA
Hint

Source

#include <iostream>
#include <string>
#include <typeinfo>
using namespace std;
typedef struct nd
{
    char data;
    nd *left, *right;
} node, *pnode;
const char gen = ',';
const int LEN = 55;
//无返回值形式，树使用引用
void createTree(pnode &root, char str[], int &index)
{
    char c;
    c = str[index];
    if(c == gen)
        root = nullptr;
    else
    {
        root = new node();
        root->left = root->right = nullptr;
        root->data = c;
        createTree(root->left, str, ++index);
        createTree(root->right, str, ++index);
    }
}
//以返回值的形式,树不使用引用
pnode createtree(char str[], int &index)
{
    char c;
    c = str[index];
    pnode root = nullptr;
    if(c != gen)
    {
        root = new node();
        root->data = c;
        root->left = createtree(str, ++index);
        root->right = createtree(str, ++index);
    }
    return root;
}
//中序遍历
void mid(node *root)
{
    if(root != nullptr)
    {
        mid(root->left);
        cout << root->data;
        mid(root->right);
    }
}
//后序遍历
void hou(node *root)
{
    if(root != nullptr)
    {
        hou(root->left);
        hou(root->right);
        cout << root->data;
    }
}
//先序中序还原二叉树
//先序遍历获得树根
//中序遍历获得左右子树
pnode xianzhong(string xian, string zhong, int xl, int xr, int zl, int zr)
{
//    cout << xl << " "
//         << xr << " "
//         << zl << " "
//         << zr << " " << endl;
    pnode root = nullptr;
    if(xl <= xr && zl <= zr)
    {
        char c = xian[xl];
        //i指向根节点
        int i = zl;
        //从中序遍历中找到树根，分开获得左右子树的字符串
        while(i < zr)
        {
            if(zhong[i] == c)
                break;
            i++;
        }
        int len = i - zl;//左子树的长度
        root = new node();
        root->data = c;
        root->left = xianzhong(xian, zhong, xl+1, xl + len, zl, i-1);
        root->right = xianzhong(xian, zhong, xl + len +1, xr, i+1, zr);
    }


    return root;
}
int main()
{
    pnode root = nullptr;
    string strx, strz;
    cin >> strx >> strz;

    int lenx = strx.length();
    root = xianzhong(strx, strz, 0, lenx-1, 0, lenx-1);
    hou(root);
    cout << endl;

    return 0;
}