T5- rebuild binary tree

Title Description

And enter the result in a preorder traversal of a binary tree in preorder traversal of the binary tree a rebuild. Suppose Results preorder traversal order and input of duplicate numbers are free. Before entering e.g. preorder traversal sequence {1,2,4,7,3,5,6,8} and {4,7,2,1,5,3,8,6} order traversal sequence, and the reconstructed binary tree return.

Time limit: C / C ++ 1 second, 2 seconds languages other
space restrictions: C / C ++ 32M, 64M other languages

code

This question is a common operation in the binary tree, the key is found in the preamble, the order of the characteristics of the function of problem-solving ideas, you can easily find the need to achieve a recursion.

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        int lenth=pre.size();//记录数组长度
        int mid=0;//记录递归过程的中间节点
        vector<int> pre_left,pre_right,vin_left,vin_right;//记录递归过程中数组的左右子数组
        
        //递归出口
        if (lenth==0)
            return NULL;
        
        //创建根节点
        TreeNode *head=new TreeNode(pre[0]);
        
        //在中序数组找根节点
        for(int i=0;i<lenth;i++)
        {
            if(vin[i]==pre[0])
            {
                mid=i;
                break;
            }
        }
        
        //前序第一个节点为根节点，对应中序中的第mid个节点
        //中序数组pre中，mid左边的数作为左子树节点中序集合，右边的数作为右子树节点中序集合
        //谦虚数组vin中，第2到（1+mid）的数作为左子树节点先序集合，剩余的数作为右子树节点先序集合
        for(int i=0;i<mid;i++)
        {
            vin_left.push_back(vin[i]);
            pre_left.push_back(pre[i+1]);
        }
        for(int i=mid+1;i<lenth;i++)
        {
            vin_right.push_back(vin[i]);
            pre_right.push_back(pre[i]);
        }
        
        //递归，递归构建左、右子树，直到叶节点
        head->left=reConstructBinaryTree(pre_left,vin_left);
        head->right=reConstructBinaryTree(pre_right,vin_right);
        return head;
    }
};