重建二叉树

根据中序遍历和前序遍历重建二叉树，同时验证是否合法。

#include <iostream>#include <string>using namespace std;#define TREELEN 6struct Node{Node* pLeft;Node* pRight;char value;};Node* rebuild(char *preOrder, char *inOrder, int nTreeLen){Node *pTemp = new Node;pTemp->value = *preOrder;pTemp->pLeft = NULL;pTemp->pRight = NULL;cout<<pTemp->value<<endl;int letter[26];for(int i=0; i<26; ++i){letter[i] = 0;}// 验证前序遍历和中序遍历字符种类个数是否一致for(int i=0; i<nTreeLen; ++i){char temp = *(preOrder+i);if(islower(temp)){letter[temp-'a']++;}else if(isupper(temp)){letter[temp-'A']++;}temp = *(inOrder+i);if(islower(temp)){letter[temp-'a']--;}else if(isupper(temp)){letter[temp-'A']--;}}for(int i=0; i<26; ++i){if(letter[i] != 0){cout<<"error string!"<<endl;throw runtime_error("error string!");}}if(nTreeLen == 1){return pTemp;}// 求左右子树长度int leftTreeLen = 0;int rightTreeLen = 0;for(int i=0; i<nTreeLen; ++i){if( *preOrder == *(inOrder+i) ){leftTreeLen = i;rightTreeLen = nTreeLen - i - 1;break;}}if(leftTreeLen > 0){pTemp->pLeft = rebuild(preOrder+1, inOrder, leftTreeLen);}if(rightTreeLen > 0){pTemp->pRight = rebuild(preOrder+leftTreeLen+1, inOrder+leftTreeLen+1, rightTreeLen);}return pTemp;}void traversal(Node* root, int flag){if(root == NULL)return;if(flag == 0){cout<<root->value<<"\t";traversal(root->pLeft, flag);traversal(root->pRight, flag);}else if(flag == 1){traversal(root->pLeft, flag);cout<<root->value<<"\t";traversal(root->pRight, flag);}}int main(){char preOrder[TREELEN] = {'a','b','d','c','e','f'};char inorder[TREELEN] = {'d','b','a','e','c','f'};Node* root = NULL;root = rebuild(preOrder, inorder, TREELEN);cout<<"\n前序遍历："<<endl;traversal(root, 0);cout<<"\n中序遍历："<<endl;traversal(root, 1);}