题解 | #重建二叉树#C++暴力哈希表递归解法

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param preOrder int整型vector 
     * @param vinOrder int整型vector 
     * @return TreeNode类
     */
    unordered_map<int,int> hash;
    TreeNode*answer(int left1,int right1,int left2,int right2,vector <int> preOrder,vector <int> vinOrder)
    {
        if(left1<=right1  &&  left2<=right2)
        {
            int m=preOrder[left1];
            int n=hash[m];
            int len2=n-left2;
            TreeNode*root=new TreeNode(m);
            root->left=answer(left1+1,left1+len2,left2,n-1,preOrder,vinOrder);
            root->right=answer(left1+len2+1,right1,n+1,right2,preOrder,vinOrder);
            return root;
        }
        return NULL;
    }
    TreeNode* reConstructBinaryTree(vector<int>& preOrder, vector<int>& vinOrder) 
    {
        int i=0;
        int len=preOrder.size();
        for(i=0;i<len;i++)
        {
            hash[vinOrder[i]]=i;
        }
        i=0;
        return answer(i,len-1,i,len-1,preOrder,vinOrder);
    }
};