题解 | #输出二叉树的右视图#

先递归建树，然后再层次遍历，每次取每层中最右边的数

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 xianxu int整型vector 先序遍历
     * @param zhongxu int整型vector 中序遍历
     * @return int整型vector
     */
    TreeNode* construct(vector<int> pre,int l1,int r1,vector<int> mid,int l2,int r2){
        if(l1>r1||l2>r2) return NULL;
        TreeNode* root=new TreeNode(pre[l1]);
        int index=l2;
        for(index;index<=r2;index++){
            if(mid[index]==pre[l1]) break;
        }
        root->left=construct(pre,l1+1,l1+index-l2,mid,l2,index-1);
        root->right=construct(pre,l1+index-l2+1,r1,mid,index+1,r2);
        return root;
    }
    vector<int> solve(vector<int>& xianxu, vector<int>& zhongxu) {
        // write code here
        TreeNode* root=construct(xianxu,0,xianxu.size()-1,zhongxu,0,zhongxu.size()-1);
        vector<int> res;
        if(root==NULL) return res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            int size=q.size();
            bool flag=true;
            while(size--){
                TreeNode* node=q.front();
                q.pop();
                if(flag){
                    res.push_back(node->val);
                    flag=false;
                }
                if(node->right) q.push(node->right);
                if(node->left) q.push(node->left);
            }
                       
        }
        return res;
    }
};