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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.
In this post, you will find the solution for the Binary Tree Right Side View in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.
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Link for the Problem – Binary Tree Right Side View– LeetCode Problem
Binary Tree Right Side View – LeetCode Problem
Problem:
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]
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
Binary Tree Right Side View– LeetCode Solutions
Binary Tree Right Side View Solution in C++:
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
if (!root)
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;
}
};
Binary Tree Right Side View Solution in Java:
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> ans = new ArrayList<>();
dfs(root, 0, ans);
return ans;
}
private void dfs(TreeNode root, int depth, List<Integer> ans) {
if (root == null)
return;
if (depth == ans.size())
ans.add(root.val);
dfs(root.right, depth + 1, ans);
dfs(root.left, depth + 1, ans);
}
}
Binary Tree Right Side View Solution in Python:
class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
ans = []
def dfs(root: Optional[TreeNode], depth: int) -> None:
if not root:
return
if depth == len(ans):
ans.append(root.val)
dfs(root.right, depth + 1)
dfs(root.left, depth + 1)
dfs(root, 0)
return ans