Binary Tree Right Side View
To solve this problem, we can perform a levelorder traversal (breadthfirst search) on the tree and keep track of the last element at each level. We will then return these last elements as the rightside view of the binary tree.
Python code:


Python code that demonstrates this approach:


Explanation:
 Empty Tree Check: If the root is
None
, we return an empty list.  Initialize Variables:
result
is initialized to hold the rightside view. Aqueue
is used to store the nodes at the current level of traversal.  BreadthFirst Traversal: Using a loop, we traverse each level of the tree. We determine the size of the current level and iterate over that many elements in the queue, processing each node and adding its children to the queue.
 Add RightSide Node: For the last node of each level (i.e.,
i == level_size  1
), we append its value to the result.  Process Children: If a node has a left or right child, we add them to the queue for processing in the next level.
The final result
list will contain the values of the nodes that can be seen from the right side of the tree.
The time complexity of this solution is (O(n)), where (n) is the number of nodes in the tree, and the space complexity is (O(n)) as well, due to the use of the queue to store nodes at the current level of traversal.
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return result
Explanation:
 Empty Tree Check: If the root is
None
, we return an empty list.  Initialize Variables:
result
is initialized to hold the rightside view. Aqueue
is used to store the nodes at the current level of traversal.  BreadthFirst Traversal: Using a loop, we traverse each level of the tree. We determine the size of the current level and iterate over that many elements in the queue, processing each node and adding its children to the queue.
 Add RightSide Node: For the last node of each level (i.e.,
i == level_size  1
), we append its value to the result.  Process Children: If a node has a left or right child, we add them to the queue for processing in the next level.
The final result
list will contain the values of the nodes that can be seen from the right side of the tree.
The time complexity of this solution is (O(n)), where (n) is the number of nodes in the tree, and the space complexity is (O(n)) as well, due to the use of the queue to store nodes at the current level of traversal.