根據(jù)前序和中序重構(gòu)二叉樹

1、#include<iostream>#include<vector>using namespace std;struct TreeNode int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) ;class Solution public:struct TreeNode* reConstructBinaryTree(vector<int> pre, vector<int> in) vector<int>:i

2、terator ibegin, iend,begin,end;ibegin = pre.begin();iend = pre.end();begin = in.begin();end = in.end();struct TreeNode *newnode = new struct TreeNode(*ibegin);vector<int>newpre1;vector<int>newpre2;vector<int>newin1;vector<int>newin2;int count = 0;for (; begin != end; )if (*be

3、gin= *ibegin)begin+;break;newin1.push_back(*begin);begin+;count+;for (; begin != end; begin+)newin2.push_back(*begin);ibegin+;for (int i=0; ibegin != iend;i+)if (i <count)newpre1.push_back(*ibegin);ibegin+;elsenewpre2.push_back(*ibegin);ibegin+;if (!newpre2.empty() newnode->right = reConstruct

4、BinaryTree(newpre2, newin2);if (!newpre1.empty() newnode->left = reConstructBinaryTree(newpre1, newin1);return newnode;void show(struct TreeNode *mytreenode)if (mytreenode = NULL)return;show(mytreenode->left);show(mytreenode->right);cout << mytreenode->val << " "int

5、 main()vector<int>myvector1;vector<int>myvector2;myvector1.push_back(1);myvector1.push_back(2);myvector1.push_back(4);myvector1.push_back(7);myvector1.push_back(3);myvector1.push_back(5);myvector1.push_back(6);myvector1.push_back(8);myvector2.push_back(4);myvector2.push_back(7);myvector2.push_back(2);myvector2.push_back(1);myvector2.push_back(5);myvector2.push_back(3);myvector2.push_back(6);myvector2.push_back(8);Solution mys;struct TreeNode *p=mys.reCo