题目：
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:You may assume that duplicates do not exist in the tree.
思路：根据前序和中序的特点 递归调用创建二叉树
#include <iostream>#include <vector>using namespace std;/**由一个树的前序和中序列来构造树 */typedef struct tree_node Tree;struct tree_node{Tree* left;Tree* right;int value;}; Tree* helper(vector<int>& pre,int pre_begin,int pre_end,vector<int>& inorder,int in_begin,int in_end){Tree* root=NULL;int mid;int i; if(pre_end -pre_begin < 0){return NULL;}else{for(i=in_begin;i<=in_end;i++)if(inorder[i] == pre[pre_begin])break;if(i>in_end)return NULL;root = new Tree;root->value = pre[pre_begin];root->left = helper(pre,pre_begin+1,pre_begin+1+i-1-in_begin,inorder,in_begin,i-1);root->right = helper(pre,pre_begin+1+i-1-in_begin+1,pre_end,inorder,i+1,in_end);return root;}}Tree* createBintree(vector<int>& pre,vector<int>& inorder){if(pre.size()==0 || inorder.size()==0)return NULL;return helper(pre,0,pre.size()-1,inorder,0,inorder.size()-1);}/*前序遍历*/ void print_pre(Tree* tree){if(tree != NULL){cout<<tree->value<<" "<<endl;print_pre(tree->left);print_pre(tree->right);}}/*中序遍历*/void print_inorder(Tree* tree){if(tree != NULL){print_inorder(tree->left);cout<<tree->value<<endl;print_inorder(tree->right);}}void print_post(Tree* root){if(root != NULL){print_post(root->left);print_post(root->right);cout<<root->value<<endl;}}int main(){int array[]={1,2,3,4,5,6,7,8};vector<int> inorder(array,array+sizeof(array)/sizeof(int));int array1[] ={4,2,1,3,7,6,5,8};vector<int> preorder(array1,array1+sizeof(array1)/sizeof(int));int array2[]={1,3,2,5,6,8,7,4};vector<int> postorder(array2,array2+sizeof(array2)/sizeof(int));Tree* root = createBintree(preorder,inorder);print_inorder(root);cout<<"============="<<endl;print_pre(root); cout<<endl;cout<<"============="<<endl;//Tree* second = createBinTree(inorder,postorder);print_post(second);return 0;}
，不要惧怕黑暗，人间没有永恒的夜晚;不要担心严寒，
