LC 199 - Binary Tree Right Side View

Question

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

Links

Question here and solution here

Solution

concept

Use a standard BFS but only keep the most right element for each level

code

# 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 []
            
        q = deque([root])
        ans = []

        while q:
            tmp = []
            for _ in range(len(q)):
                node = q.popleft()
                tmp.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(tmp[-1])

        return ans
        
class NeetSolution:
    """
    time complexity: O(n)
    space complexity: O(n)
    """
    def rightSideView(self, root: Optional[TreeNode]) -> List[int]:
        res = []
        q = deque([root])

        while q:
            rightSide = None # if root is None, return []
            qLen = len(q)

            for i in range(qLen):
                node = q.popleft()
                if node:
                    rightSide = node
                    q.append(node.left)
                    q.append(node.right)
            # q is non-empty, but rightSide is could be None
            if rightSide:
                res.append(rightSide.val)
        return res

Complexity

time: $O(n)$
space: $O(n)$