Binary Tree Right Side View

Clarify the problem:

• The problem requires us to return the right side view of a binary tree.
• The right side view consists of all the nodes that are visible when looking at the tree from the right side.
• We need to return the values of these visible nodes in the order from top to bottom.

Analyze the problem:

• Input: The root node of a binary tree.
• Output: A list of values representing the right side view of the tree.
• Constraints:
  • The number of nodes in the tree is in the range [0, 100].
  • Each node's value is a unique integer.

Design an algorithm:

• We can solve this problem using a depth-first search (DFS) traversal of the binary tree.
• During the traversal, we track the maximum depth reached so far and the corresponding value of the rightmost node at each depth.
• We start the traversal from the root node, and at each level, we visit the right subtree first before the left subtree.
• While traversing the tree, if the current depth is greater than the maximum depth reached so far, we add the value of the current node to the result list.
• After completing the traversal, we will have the right side view of the binary tree stored in the result list.

Explain your approach:

• We will implement a recursive function that performs the depth-first search traversal.
• During the traversal, we will keep track of the maximum depth and the corresponding value of the rightmost node at each depth.
• We will start the traversal from the root node and explore the right subtree first before the left subtree.
• If the current depth is greater than the maximum depth reached so far, we will add the value of the current node to the result list.
• Finally, we will return the result list as the right side view of the binary tree.

Write clean and readable code:

python

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right


def rightSideView(root):
    def dfs(node, depth, result):
        if not node:
            return
        if depth >= len(result):
            result.append(node.val)
        dfs(node.right, depth + 1, result)
        dfs(node.left, depth + 1, result)

    result = []
    dfs(root, 0, result)
    return result

Test your code:

• We will test the code with different test cases, including edge cases and corner cases, to ensure its correctness.
• Test Case 1: Empty tree

  python

root = None
print(rightSideView(root))
Output: []

python

root = TreeNode(1)
print(rightSideView(root))
Output: [1]

python

• root = TreeNode(1)
  root.left = TreeNode(2)
  root.right = TreeNode(3)
  root.left.right = TreeNode(5)
  root.right.right = TreeNode(4)
  print(rightSideView(root))
  Output: [1, 3, 4]
  

Optimize if necessary:

• The current solution has a time complexity of O(N), where N is the number of nodes in the binary tree.
• This is because we perform a depth-first search traversal of all the nodes in the tree.
• Since we are not allowed to use predefined functions, the current solution is already optimal.

Handle error cases:

• The code handles the case of an empty tree by returning an empty list.
• There are no other error cases to consider in this problem.

Discuss complexity analysis:

• The time complexity of the solution is O(N), where N is the number of nodes in the binary tree.
• This is because we visit each node once during the depth-first search traversal.
• The space complexity is O(H), where H is the height of the binary tree.
• In the worst case, the tree can be linear, resulting in a space complexity of O(N).
• However, in a balanced tree, the space complexity is O(log N) due to the recursive calls on the stack.
• Overall, the solution is efficient and meets the problem requirements.