题目
https://leetcode-cn.com/problems/binary-tree-right-side-view
给定一棵二叉树,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值。
示例:
输入: [1,2,3,null,5,null,4]
输出: [1, 3, 4]
解释:
1 <---
/
2 3 <---
5 4 <---
解题思路
- BFS,按层遍历节点
- 构造多层列表,将同层节点从左到右存于同一列表
- 返回结果时,取每层列表的最后一个即可
代码
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
if not root: return []
# - keep nodes in same level in level_list[level]
level_list = []
# - (node, level)
q = [(root, 1)]
while q:
node,level = q.pop(0)
# - expand level_list for next level
while level > len(level_list):
level_list.append([])
# - append nodes from left to right
level_list[level-1].append(node.val)
level += 1
if node.left: q.append((node.left, level))
if node.right: q.append((node.right, level))
# - return only right most node of each level
return [nodes[-1] for nodes in level_list]
解题思路
- 实际上没有必要存储整层的节点,只需要不停的更新最后一个节点就好
代码
class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
if not root: return []
# - level_list[level] is the right most node val at level
level_list = []
# - (node, level)
q = [(root, 1)]
while q:
node,level = q.pop(0)
# - expand level_list for next level
while level > len(level_list):
level_list.append(node.val)
# - append nodes from left to right
level_list[level-1] = node.val
level += 1
if node.left: q.append((node.left, level))
if node.right: q.append((node.right, level))
# - return only right most node of each level
return level_list