Leetcode 199 | 程序员灯塔

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()]

程序员灯塔
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