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.
For example:
Given the following binary tree,

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

You should return [1, 3, 4].

Solution1：從右側 pre-order DFS 遞歸寫法

思路： 每層只加第一次喇伯，通過cur_level和result.length控制
Time Complexity: O(N) Space Complexity: O(N) 遞歸緩存

Solution2： BFS 從右側push

思路： 每層只加第一個剂跟。 當然從左邊開始push也一樣，只加最后
Time Complexity: O(N) Space Complexity: O(N)

Solution3： Divide and Conquer (not efficient, only for demo)

Time Complexity:
if nearly balance, T(N) = 2T(N/2) + logN => O(N) ?
worst: O(N^2)
Space Complexity: O(N)

Solution1 Code：

class Solution1 {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        rightView(root, result, 0);
        return result;
    }
    
    public void rightView(TreeNode curr, List<Integer> result, int cur_level){
        if(curr == null) return;
        
        if(cur_level == result.size()){
            result.add(curr.val);
        }
        
        rightView(curr.right, result, cur_level + 1);
        rightView(curr.left, result, cur_level + 1);
    }
}

Solution2 Code：

class Solution2 {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        if(root == null) return result;
        
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        
        queue.offer(root);
        while(!queue.isEmpty()) {
            int num_on_level = queue.size();
            for(int i = 0; i < num_on_level; i++) {
                TreeNode cur = queue.poll();
                if(i == 0) result.add(cur.val);
                if(cur.right != null) queue.add(cur.right);
                if(cur.left != null) queue.add(cur.left);
                
            }
        }
        return result;
    }
}

Solution3 Code：

class Solution3 {
    public List<Integer> rightSideView(TreeNode root) {
        if(root==null)
            return new ArrayList<Integer>();
        List<Integer> left = rightSideView(root.left);
        List<Integer> right = rightSideView(root.right);
        List<Integer> re = new ArrayList<Integer>();
        re.add(root.val);
        for(int i=0;i < Math.max(left.size(), right.size()); i++){
            if(i >= right.size())
                re.add(left.get(i));
            else
                re.add(right.get(i));
        }
        return re;
    }
}