# 199. Binary Tree Right Side View

#### Medium

***

Given the `root` of 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*.

 

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/02/14/tree.jpg)

```
Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]
```

**Example 2:**

```
Input: root = [1,null,3]
Output: [1,3]
```

**Example 3:**

```
Input: root = []
Output: []
```

 

**Constraints:**

* The number of nodes in the tree is in the range `[0, 100]`.
* `-100 <= Node.val <= 100`

```python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        q = [root, "$"]
        result = []
        prev = None
        while q:
            node = q.pop(0)
            if node == "$":
                if prev is not None:
                    result.append(prev.val)
                if len(q) > 0:
                    prev = None
                    q.append("$")
                continue
            if node.left:
                q.append(node.left)
            if node.right:
                q.append(node.right)
            prev = node
        return result
```

#### Print Left View with Slight Modification

```python
class Solution:
    def leftSideView(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        q = [root, "$"]
        result = []
        prev = None
        newLevel = True
        while q:
            node = q.pop(0)
            if node == "$":
                if prev is not None:
                    pass
                    # result.append(prev.val)
                if len(q) > 0:
                    prev = None
                    newLevel = True
                    q.append("$")
                continue
            if newLevel:
                newLevel = False
                result.append(node.val)
            if node.left:
                q.append(node.left)
            if node.right:
                q.append(node.right)
            prev = node
        return result
```