[leetcode] 199. Binary Tree Right Side View @ python

最新推荐文章于 2022-09-24 13:12:34 发布

原创最新推荐文章于 2022-09-24 13:12:34 发布 · 370 阅读

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本文介绍了一种解决二叉树右视图问题的算法，通过广度优先搜索（BFS）遍历二叉树，收集从顶部到底部的可见节点值。此算法的时间复杂度为O(n)，空间复杂度为O(1)，适用于任何规模的二叉树。

原题

Given 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:

Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:

1 <—
/
2 3 <—
\
5 4 <—

解法

BFS. 按层将二叉树的值放入列表, 然后取每层列表的最后一个值即可.
Time: O(n), n为二叉树的节点数
Space: O(1)

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):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        res = []
        # base case
        if not root: return res
        # bfs
        q = [root]
        while q:
            res.append([node.val for node in q])
            new_q = []
            for node in q:
                if node.left:
                    new_q.append(node.left)
                if node.right:
                    new_q.append(node.right)
            q = new_q
            
        res = [l[-1] for l in res]
        return res