What is the key difference between BFS and DFS in solving the Binary Tree Right Side View problem?

BFS processes nodes level by level, ensuring the last node at each level is captured, while DFS explores the tree depth-first, preferring right-side nodes.

How do I handle null nodes or missing children in the Binary Tree Right Side View?

Ensure you handle null nodes appropriately by checking for them in your traversal process. In BFS, these are naturally skipped; in DFS, ensure recursion handles them correctly.

Can I solve the Binary Tree Right Side View problem with just an iterative DFS?

Yes, an iterative DFS can be used, typically with a stack, by tracking the depth and ensuring you only capture the rightmost node at each level.

How does the tree's shape impact the solution to the Binary Tree Right Side View?

The tree's shape affects the space complexity of DFS and BFS. A more unbalanced tree will impact DFS's stack depth, while BFS requires more space to handle a wide tree.

What is the optimal way to solve the Binary Tree Right Side View if the tree is very large?

For large trees, an optimal solution would likely involve BFS to minimize recursion stack depth and avoid performance issues in unbalanced trees.

#199

Medium

auto_awesome二分·树·traversal

LeetCode 题解工作台

二叉树的右视图

给定一个二叉树的根节点 root ，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 1：输入： root = [1,2,3,null,5,null,4] 输出： [1,3,4] 解释：示例 2：输入： root = [1,2,3,4,null,null,nu…

树深度优先搜索广度优先搜索二叉树

题目描述

给定一个二叉树的 根节点 root，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 1：

输入：root = [1,2,3,null,5,null,4]

输出：[1,3,4]

解释：

示例 2：

输入：root = [1,2,3,4,null,null,null,5]

输出：[1,3,4,5]

解释：

示例 3：

输入：root = [1,null,3]

输出：[1,3]

示例 4：

输入：root = []

输出：[]

提示:

二叉树的节点个数的范围是 [0,100]
-100 <= Node.val <= 100

lightbulb

解题思路

方法一：BFS

我们可以使用广度优先搜索，定义一个队列 $\textit{q}$ ，将根节点放入队列中。每次从队列中取出当前层的所有节点，对于当前节点，我们先判断右子树是否存在，若存在则将右子树放入队列中；再判断左子树是否存在，若存在则将左子树放入队列中。这样每次取出队列中的第一个节点即为该层的右视图节点。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为二叉树节点个数。

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# 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]:
        ans = []
        if root is None:
            return ans
        q = deque([root])
        while q:
            ans.append(q[0].val)
            for _ in range(len(q)):
                node = q.popleft()
                if node.right:
                    q.append(node.right)
                if node.left:
                    q.append(node.left)
        return ans

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Can the candidate explain the difference in approach between BFS and DFS for this problem?
question_mark
Does the candidate understand the importance of tracking the rightmost nodes?
question_mark
Does the candidate optimize space by choosing the appropriate traversal strategy for the tree's shape?

warning

常见陷阱

外企场景

error
Confusing the BFS and DFS traversal methods or not choosing the optimal one for the problem's constraints.
error
Failing to account for the edge case of an empty tree or ensuring the correct handling of null nodes.
error
Incorrectly managing the state during traversal, leading to missing or extra nodes in the output.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the tree is extremely unbalanced? Can you still solve it efficiently?
arrow_right_alt
How does the solution change if you need to return the left side view instead of the right?
arrow_right_alt
Can you solve the problem without using recursion?

help

常见问题

外企场景

继续练习

#226 翻转二叉树 #117 填充每个节点的下一个右侧节点指针 II #116 填充每个节点的下一个右侧节点指针