【leetcode】199. Binary Tree Right Side View

题目：
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

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思路：
先序遍历（先右后左）。max表示已经访问的最大深度，只有比这个max更大时才会保存到ret数组（答案都放在这里）中。

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代码实现：
submit一次就过了：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void f(TreeNode *root, int i, int &max, vector<int> &ret){
        if (root == nullptr){
            return ;
        }
        if (i > max){
            max = i;
            ret.push_back(root->val);
        }
        f(root->right, i+1, max, ret);
        f(root->left, i+1, max, ret);
    }
    vector<int> rightSideView(TreeNode* root) {
        int max = -1;
        vector<int> ret;
        f(root, 0, max, ret);
        
        return ret;
    }
};