二叉树的三种先序后序中序遍历的互相推出（例UVA 536

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关于树的先序中序后序遍历

由先序中序推后序

因为先序：根左右

中序：左根右

后序：右根左

根据他们的对应关系推出树

1，先序遍历中序遍历推后序遍历：

#include <iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int maxn=50;void tree(int n,char *s1, char *s2,char *s)  //构建树{    if(n<=0)        return;    int p= strchr(s2,s1[0])-s2;  // 在中序遍历中寻找根的位置    tree(p,s1+1,s2,s);      // 生成左子树，    tree(n-p-1,s1+p+1,s2+p+1,s+p); // 生成右子树    s[n-1]=s1[0];  // 将先序的前值赋给后序的末值}int main(){    char s[maxn],s1[maxn],s2[maxn];    while(~(scanf("%s%s",&s1,&s2)))    {        int n=strlen(s1);        tree(n,s1,s2,s);        s[n]='\0';        printf("%s\n",s);    }    return 0;}

以上为 UVA 536 - Tree Recovery点击打开链接 (二叉树)ac代码

但是。。。

由后序中序去推前序崩掉了。。。。可能是因为推的不行吧

从网上看了看思路思考一番，蒙出来了

2，先序遍历中序遍历推后序遍历：

#include <iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespace std;const int maxn=50;int l;void tree(int n,char *s1, char *s2,char *s){    if(n<=0)        return;        int p=0;   for(;p<n;p++)   {       if(s2[p]==s1[n-1])        break;   }    tree(p,s1,s2,s+1);    tree(n-p-1,s1+p,s2+p+1,s+p+1);    s[0]=s1[n-1];}int main(){    char s[maxn],s1[maxn],s2[maxn];    while(~(scanf("%s%s",s2,s1)))    {       memset(s,0,sizeof(s));        int l=strlen(s1);        tree(l,s1,s2,s);        s[l]='\0';        printf("%s\n",s);    }    return 0;}

关于借鉴的大神的代码

    /***************************************************************************/                      /*已知中序、后序遍历，求前序遍历*/      /***************************************************************************/            #include <iostream>      #include <fstream>      #include <string>            using namespace std;            struct TreeNode      {          struct TreeNode* left;          struct TreeNode* right;          char  elem;      };            TreeNode* BinaryTreeFromOrderings(char* inorder, char* aftorder, int length)      {          if(length == 0)          {              return NULL;          }          TreeNode* node = new TreeNode;//Noice that [new] should be written out.          node->elem = *(aftorder+length-1);          cout<<node->elem;          int rootIndex = 0;          for(;rootIndex < length; rootIndex++)//a variation of the loop          {              if(inorder[rootIndex] ==  *(aftorder+length-1))                  break;          }          node->left = BinaryTreeFromOrderings(inorder, aftorder , rootIndex);          node->right = BinaryTreeFromOrderings(inorder + rootIndex + 1, aftorder + rootIndex , length - (rootIndex + 1));                    return node;      }            int main(int argc, char** argv)      {          printf("Question: 已知中序、后序遍历，求前序遍历\n\n");                 char* in="ADEFGHMZ";    //中序          char* af="AEFDHZMG";    //后序                cout<<"中序是："<<in<<endl;          cout<<"后序是："<<af<<endl<<endl;          cout<<"前序是：";                BinaryTreeFromOrderings(in, af, 8);           printf("\n\n");          return 0;      }

/***************************************************************************/                  /*已知前序、中序遍历，求后序遍历*/  /***************************************************************************/    #include <iostream>    #include <fstream>    #include <string>      using namespace std;    struct TreeNode  {      struct TreeNode* left;      struct TreeNode* right;      char  elem;  };    void BinaryTreeFromOrderings(char* inorder, char* preorder, int length)  {      if(length == 0)          {              //cout<<"invalid length";              return;          }      TreeNode* node = new TreeNode;//Noice that [new] should be written out.      node->elem = *preorder;      int rootIndex = 0;      for(;rootIndex < length; rootIndex++)      {          if(inorder[rootIndex] == *preorder)          break;      }      //Left      BinaryTreeFromOrderings(inorder, preorder +1, rootIndex);      //Right      BinaryTreeFromOrderings(inorder + rootIndex + 1, preorder + rootIndex + 1, length - (rootIndex + 1));      cout<<node->elem;      return;  }     int main(int argc, char* argv[])  {      printf("Question: 已知前序、中序遍历，求后序遍历\n\n");        char* pr="GDAFEMHZ";//前序      char* in="ADEFGHMZ";//中序          cout<<"前序是："<<pr<<endl;      cout<<"中序是："<<in<<endl<<endl;      cout<<"后序是：";        BinaryTreeFromOrderings(in, pr, 8);      printf("\n\n");      return 0;

来源点击打开链接

还有非递归求法（点击打开链接）

完美，以后再细看一次

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