4、根据前序和中序，重建二叉树

class Solution {
public:
    TreeNode* reConstructBinaryTree(vector<int> pre,vector<int> in) {
        int len = in.size();
        if(len == 0)
            return NULL;
        vector<int> left_pre,right_pre,left_in,right_in;
        TreeNode* head = new TreeNode(pre[0]);
        int k = 0;
        for(int i = 0;i < len;i++)
        {
            if(in[i] == pre[0])
            {
                k = i;
                break;
            }
        }
        for(int i = 0;i < k;i++)
        {
            left_in.push_back(in[i]);
            left_pre.push_back(pre[i+1]);
        }
        for(int i = k + 1;i < len;i++)
        {
            right_in.push_back(in[i]);
            right_pre.push_back(pre[i]);
        }
        head->left = reConstructBinaryTree(left_pre,left_in);
        head->right = reConstructBinaryTree(right_pre,right_in);
        return head;
    }
};