199. Binary Tree Right Side View

Medium LeetCode

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 = [1,2,3,4,null,null,null,5]

Output: [1,3,4,5]

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

How to solve the problem

Code

• Recursion (DFS)

# 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]:
        # DFS
        ans = []
        def dfs(node, depth):
            if node is None:
                return 
            if depth == len(ans):
                ans.append(node.val)
            dfs(node.right, depth+1)
            dfs(node.left, depth+1)
        dfs(root, 0)
        return ans

Complexity

• Time complexity: O(n), n == number of nodes
• Space complexity: O(h), h == height of tree