已知二叉树的中序遍历和后序遍历，如何求前序遍历

    

     在已知二叉树的前序遍历和中序遍历的情况下，求出了后序遍历。

那么，对称地，如果已知二叉树的后序遍历和中序遍历，如何求前序遍历呢？

其实思路与上文完全类似。

代码如下：

[cpp] view plaincopyprint?

1. // InPost2Pre.cpp : Defines the entry point for the console application. 
2. // 
3.  
4. #include "stdafx.h" 
5. #include <iostream> 
6. using namespace std; 
7.  
8. struct TreeNode 
9. { 
10.   struct TreeNode* left; 
11.   struct TreeNode* right; 
12.   char  elem; 
13. }; 
14.  
15. TreeNode* BinaryTreeFromOrderings(char* inorder,char* postorder,int length) 
16. { 
17.   if(length == 0) 
18.     { 
19.       return NULL; 
20.     } 
21.   //We have root then; 
22.   TreeNode* node = new TreeNode;//Noice that [new] should be written out. 
23.   node->elem = *(postorder+length-1); 
24.   cout<<node->elem<<endl; 
25.   int rootIndex = 0; 
26.   for(;rootIndex <length; rootIndex++) 
27.     { 
28.       if(inorder[rootIndex] == *(postorder+length-1)) 
29.       break; 
30.     } 
31.   //在循环条件而非循环体完成了对变量值的改变。 
32.   //Left 
33.   node->left = BinaryTreeFromOrderings(inorder, postorder, rootIndex); 
34.   //Right 
35.   node->right = BinaryTreeFromOrderings(inorder + rootIndex + 1, postorder + rootIndex, length - (rootIndex+1)); 
36.   //node->right = *(postorder+length-2); 
37.  
38.   return node; 
39. } 
40.  
41.  
42. int main(int argc,char* argv[]) 
43. { 
44.     printf("Hello World!\n"); 
45.     char* in="ADEFGHMZ"; 
46.     char* po="AEFDHZMG";     
47.         BinaryTreeFromOrderings(in, po, 8); 
48.         printf("\n"); 
49.     return 0; 
50. } 

// InPost2Pre.cpp : Defines the entry point for the console application.//#include "stdafx.h"#include <iostream>using namespace std;struct TreeNode{  struct TreeNode* left;  struct TreeNode* right;  char  elem;};TreeNode* BinaryTreeFromOrderings(char* inorder, char* postorder, int length){  if(length == 0)    {      return NULL;    }  //We have root then;  TreeNode* node = new TreeNode;//Noice that [new] should be written out.  node->elem = *(postorder+length-1);  cout<<node->elem<<endl;  int rootIndex = 0;  for(;rootIndex <length; rootIndex++)    {      if(inorder[rootIndex] == *(postorder+length-1))      break;    }  //在循环条件而非循环体完成了对变量值的改变。  //Left  node->left = BinaryTreeFromOrderings(inorder, postorder, rootIndex);  //Right  node->right = BinaryTreeFromOrderings(inorder + rootIndex + 1, postorder + rootIndex, length - (rootIndex+1));  //node->right = *(postorder+length-2);  return node;}int main(int argc, char* argv[]){printf("Hello World!\n");char* in="ADEFGHMZ";char* po="AEFDHZMG";        BinaryTreeFromOrderings(in, po, 8);        printf("\n");return 0;}