https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6?tpId=265&tqId=39211&rp=1&ru=/exam/oj/ta&qru=/exam/oj/ta&sourceUrl=%2Fexam%2Foj%2Fta%3FtpId%3D13&difficulty=undefined&judgeStatus=undefined&tags=&title=

/**
 * 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* BinaryTree(vector<int> &pre, int preBegin, int preEnd, vector<int> &vin, int vinBegin, int vinEnd)
    {
        if((preBegin>preEnd) || (vinBegin>vinEnd))
        {
            return nullptr;
        }

        TreeNode* root = new TreeNode(pre[preBegin]);
        int rooti=pre[preBegin];
        for(int i=vinBegin; i<=vinEnd; ++i)
        {
            if(vin[i]==rooti)    //  找到根结点下标
            {
                root->left = BinaryTree(pre, preBegin+1, i-vinBegin+preBegin, vin, vinBegin, i-1);
                root->right = BinaryTree(pre, i-vinBegin+preBegin+1, preEnd, vin, i+1, vinEnd);
                break;
            }
        }

        return root;
    }

    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> vin) {
        if(pre.size() != vin.size())
        {
            return nullptr;
        }

        return BinaryTree(pre, 0, pre.size()-1, vin, 0, vin.size()-1);
    }
};