Leetcode 199

标签：right 199 val self rightmost Leetcode left

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()]

标签：right,199,val,self,rightmost,Leetcode,left
From： https://www.cnblogs.com/selfmade-Henderson/p/17009241.html