199. 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.
Analysis: given the depth, each depth we can only see the rightmost node(leaf) in exists. So actually the question is asking for a DFS for the rightmost nodes each stage.
So we can have a BFS, keep each level's rightmost node in a queue. Then output the nodes in the queue.
# 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]:
d = {}
def f(r,i):
if r:
d[i] = r.val
f(r->right, i+1)
f(r->left, i+1)
f(root,0)
return d[*values()]
