题解 | #输出二叉树的右视图#C++先构造，再层次遍历

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
     *
     * 求二叉树的右视图
     * @param preOrder int整型vector 先序遍历
     * @param inOrder int整型vector 中序遍历
     * @return int整型vector
     */
    unordered_map<int,int> hash;
    TreeNode*answer1(int left1,int right1,int left2,int right2,vector <int> preOrder,vector <int> inOrder)
    {
        if(left1<=right1  &&  left2<=right2)
        {
            TreeNode*ans=new TreeNode(preOrder[left1]);
            int m=hash[preOrder[left1]];
            int n=m-left2;
            ans->left=answer1(left1+1,left1+n,left2,m-1,preOrder,inOrder);
            ans->right=answer1(left1+n+1,right1,m+1,right2,preOrder,inOrder);
            return ans;
        }
        return NULL;
    }
    int height(TreeNode*root)
    {
        if(root)
        {
            int m=height(root->left);
            int n=height(root->right);
            return max(m,n)+1;
        }
        return 0;
    }
    void answer3(TreeNode*root,int h,vector <vector <int>>&ans)
    {
        if(root)
        {
            answer3(root->left,h+1,ans);
            ans[h].push_back(root->val);
            answer3(root->right,h+1,ans);
        }
    }
    vector<int> solve(vector<int>& preOrder, vector<int>& inOrder) 
    {
        int i=0;
        int len=preOrder.size();
        for(i=0;i<len;i++)
        {
            hash[inOrder[i]]=i;
        }
        TreeNode* ans2=answer1(0,len-1,0,len-1,preOrder,inOrder);
        int h=height(ans2);
        vector <vector <int>>ans(h);
        answer3(ans2,0,ans);
        vector <int> z;
        for(i=0;i<h;i++)
        {
            int len3=ans[i].size();
            z.push_back(ans[i][len3-1]);
        }
        return z;
    }
};