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2

输入: [1,2,3,null,5,null,4]
输出: [1,3,4]

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2

输入: [1,null,3]
输出: [1,3]

1
2

输入: []
输出: []

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// @lc code=start
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    /// ## 解题思路
    /// - 层历，取每一层的最后一个元素
    pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut right_view_vals = Vec::new();
        let mut q = VecDeque::new();
        if let Some(root) = root {
            q.push_back(root.clone());
            while !q.is_empty() {
                let mut cur_vals = Vec::new();
                let mut next_q = VecDeque::new();
                while let Some(node) = q.pop_front() {
                    cur_vals.push(node.borrow().val);
                    if let Some(left) = &node.borrow().left {
                        next_q.push_back(left.clone());
                    }
                    if let Some(right) = &node.borrow().right {
                        next_q.push_back(right.clone());
                    }
                }
                right_view_vals.push(*cur_vals.last().unwrap());
                if !next_q.is_empty() {
                    q = next_q;
                }
            }
        }

        right_view_vals
    }
}
// @lc code=end