199. Binary Tree Right Side View LeetCode Solution

In this guide, you will get 199. Binary Tree Right Side View LeetCode Solution with the best time and space complexity. The solution to Binary Tree Right Side View problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Binary Tree Right Side View solution in C++
Binary Tree Right Side View solution in Java
Binary Tree Right Side View solution in Python
Additional Resources

199. Binary Tree Right Side View LeetCode Solution image

Problem Statement of Binary Tree Right Side View

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:

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

Example 2:

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

Example 3:

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

Example 4:

Input: root = []
Output: []

Constraints:

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

199. Binary Tree Right Side View LeetCode Solution in C++

class Solution {
 public:
  vector<int> rightSideView(TreeNode* root) {
    if (root == nullptr)
      return {};

    vector<int> ans;
    queue<TreeNode*> q{{root}};

    while (!q.empty()) {
      const int size = q.size();
      for (int i = 0; i < size; ++i) {
        TreeNode* node = q.front();
        q.pop();
        if (i == size - 1)
          ans.push_back(node->val);
        if (node->left)
          q.push(node->left);
        if (node->right)
          q.push(node->right);
      }
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

199. Binary Tree Right Side View LeetCode Solution in Java

class Solution {
  public List<Integer> rightSideView(TreeNode root) {
    if (root == null)
      return new ArrayList<>();

    List<Integer> ans = new ArrayList<>();
    Queue<TreeNode> q = new ArrayDeque<>(List.of(root));

    while (!q.isEmpty()) {
      final int size = q.size();
      for (int i = 0; i < size; ++i) {
        TreeNode node = q.poll();
        if (i == size - 1)
          ans.add(node.val);
        if (node.left != null)
          q.offer(node.left);
        if (node.right != null)
          q.offer(node.right);
      }
    }

    return ans;
  }
}
// code provided by PROGIEZ

199. Binary Tree Right Side View LeetCode Solution in Python

class Solution:
  def rightSideView(self, root: TreeNode | None) -> list[int]:
    if not root:
      return []

    ans = []
    q = collections.deque([root])

    while q:
      size = len(q)
      for i in range(size):
        root = q.popleft()
        if i == size - 1:
          ans.append(root.val)
        if root.left:
          q.append(root.left)
        if root.right:
          q.append(root.right)

    return ans
 # code by PROGIEZ