剑指offer 重建二叉树

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/**  * Definition for binary tree  * 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) {         if(pre.empty())         return NULL;         if(pre.size()==1)             return new TreeNode(pre[0]);            TreeNode *root=new TreeNode(pre[0]);         vector<int>::iterator ite1=in.begin();         vector<int>::iterator ite2=in.end();         int leftNum=0;         int rightNum=0;         for(;ite1!=ite2;ite1++)         {               if(*ite1!=root->val)leftNum++;             else break;         }         rightNum=pre.size()-1-leftNum;         vector<int>lson_pre(leftNum,0);         lson_pre.assign(pre.begin()+1,pre.begin()+1+leftNum);         vector<int>lson_in(leftNum,0);         lson_in.assign(in.begin(),in.begin()+leftNum);         vector<int>rson_pre(rightNum,0);         rson_pre.assign(pre.end()-rightNum,pre.end());         vector<int>rson_in(rightNum,0);         rson_in.assign(in.end()-rightNum,in.end());            root->left=reConstructBinaryTree(lson_pre,lson_in);         root->right=reConstructBinaryTree(rson_pre,rson_in);               return root;     } };