199. Binary Tree Right Side View

Python Solution

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        
        result = []
        queue = deque([root])
        
        while queue:
            level_size = len(queue)
            
            for i in range(level_size):
                node = queue.popleft()
                
                # If it's the last node in the level
                if i == level_size - 1:
                    result.append(node.val)
                
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
        
        return result

DFS Solution:

class Solution:
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        def dfs(node: TreeNode, level: int, result: List[int]) -> None:
            if not node:
                return
            
            # Add the first node we see at this level
            if level == len(result):
                result.append(node.val)
            
            # Visit right first to get right side view
            dfs(node.right, level + 1, result)
            dfs(node.left, level + 1, result)
        
        result = []
        dfs(root, 0, result)
        return result

Java Solution

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public List rightSideView(TreeNode root) {
        List result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        
        Queue queue = new LinkedList<>();
        queue.offer(root);
        
        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.poll();
                
                // If it's the last node in the level
                if (i == levelSize - 1) {
                    result.add(node.val);
                }
                
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
        }
        
        return result;
    }
}

C++ Solution

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector rightSideView(TreeNode* root) {
        vector result;
        if (!root) {
            return result;
        }
        
        queue q;
        q.push(root);
        
        while (!q.empty()) {
            int levelSize = q.size();
            
            for (int i = 0; i < levelSize; i++) {
                TreeNode* node = q.front();
                q.pop();
                
                // If it's the last node in the level
                if (i == levelSize - 1) {
                    result.push_back(node->val);
                }
                
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
        }
        
        return result;
    }
};

JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var rightSideView = function(root) {
    if (!root) {
        return [];
    }
    
    const result = [];
    const queue = [root];
    
    while (queue.length) {
        const levelSize = queue.length;
        
        for (let i = 0; i < levelSize; i++) {
            const node = queue.shift();
            
            // If it's the last node in the level
            if (i === levelSize - 1) {
                result.push(node.val);
            }
            
            if (node.left) {
                queue.push(node.left);
            }
            if (node.right) {
                queue.push(node.right);
            }
        }
    }
    
    return result;
};

C# Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public IList RightSideView(TreeNode root) {
        IList result = new List();
        if (root == null) {
            return result;
        }
        
        Queue queue = new Queue();
        queue.Enqueue(root);
        
        while (queue.Count > 0) {
            int levelSize = queue.Count;
            
            for (int i = 0; i < levelSize; i++) {
                TreeNode node = queue.Dequeue();
                
                // If it's the last node in the level
                if (i == levelSize - 1) {
                    result.Add(node.val);
                }
                
                if (node.left != null) {
                    queue.Enqueue(node.left);
                }
                if (node.right != null) {
                    queue.Enqueue(node.right);
                }
            }
        }
        
        return result;
    }
}