# 199. Binary Tree Right Side View

### Description

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.

### Constraints

### Approach

### Links

* [GeeksforGeeks](https://www.geeksforgeeks.org/print-right-view-binary-tree-2/)
* [Leetcode](https://leetcode.com/problems/binary-tree-right-side-view/)
* [ProgramCreek](https://www.programcreek.com/2014/04/leetcode-binary-tree-right-side-view-java/)
* YouTube

### **Examples**

{% tabs %}
{% tab title="Example 1" %}
**Input:** \[1, 2, 3, null, 5, null, 4]

**Output:** \[1, 3, 4]

**Explanation:**

<div align="left"><img src="/files/-MJ5mP69GvKQwiCyQ6_C" alt=""></div>
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="TreeNode" %}

```java
// Definition for a binary tree node.
public class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;
	TreeNode() {}
	
	TreeNode(int val) { 
		this.val = val;
	}
	
	TreeNode(int val, TreeNode left, TreeNode right) {
		this.val = val;
		this.left = left;
		this.right = right;
	}
}
```

{% endtab %}

{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(N) since one has to visit each node.
 * Space complexity : O(D) to keep the queues, where D is a tree diameter. 
 *    Let's use the last level to estimate the queue size. This level could 
 *    contain up to N/2 tree nodes in the case of complete binary tree.
 */

class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> result = new ArrayList();
        
        if(root == null) return result;
        
        LinkedList<TreeNode> queue = new LinkedList();
        queue.add(root);
        
        while(!queue.isEmpty()) {
            int size = queue.size();
            for(int i = 0; i < size; i++) {
                TreeNode node = queue.remove();
                if(i == 0) {
                    result.add(node.val);
                }
                if(node.right != null) {
                    queue.add(node.right);
                }
                if(node.left != null) {
                    queue.add(node.left);
                }
            }
        }
        return result;
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : O(N) since one has to visit each node.
 * Space complexity : O(H) to keep the recursion stack, where H is a tree height. 
 *    The worst-case situation is a skewed tree, when H=N.
 */
 
 class Solution {
    List<Integer> rightside = new ArrayList();
    
    public List<Integer> rightSideView(TreeNode root) {
        if (root == null) return rightside;
        
        helper(root, 0);
        return rightside;
    }
    
    public void helper(TreeNode node, int level) {
        if (level == rightside.size()) {
            rightside.add(node.val);
        }
        if (node.right != null) {
            helper(node.right, level + 1);
        }
        if (node.left != null) {
            helper(node.left, level + 1);
        }
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*


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