其实题目的意思就是相当于二叉树每一行的最右边的一个元素,那么简单,先bfs后取每一个数组的最后一位数就可以了,代码如下: 1/**2* Definition for a binary tree node.3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NU...
TreeNode*node =q.front(); q.pop();if(node->left) q.push(node->left);if(node->right) q.push(node->right); } }returnres; } }; LeetCode All in One 题目讲解汇总(持续更新中...)
问题链接 英文网站:199. Binary Tree Right Side View中文网站:199. 二叉树的右视图问题描述Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the n…
# Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # 黄哥Python培训 黄哥所写 class Solution: def rightSideView(self, root: TreeNode) -> List[int]: if root is None: return [] res = [...
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 <--- ...
【LeetCode】199. Binary Tree Right Side View 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/binary-tree-right-side-view/description/ 题目描述: Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered...
http://www.programcreek.com/2014/04/leetcode-binary-tree-right-side-view-java/ 看了这篇博客就懂了。 ** 总结: queue来遍历tree ** Anyway, Good luck, Richardo! My code: /** * Definition for a binary tree node. * public class TreeNode { ...
图片来源:https://leetcode.com/articles/binary-tree-right-side-view/ 对于时间复杂度来说,基本上都是O(n),因为要访问所有的点。 对于空间复杂度来说,BFS取决于扫描过程中每层的node数,就是树的宽度,而DFS取决于扫描过程中树的深度。最坏情况两个都是O(n)。
Leetcode 226. Invert Binary Tree \ to 代码语言:javascript 代码运行次数:0 4/\72/\/\9631 Trivia: This problem was inspired bythis original tweetbyMax Howell: Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck...
Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 栈迭代 复杂度 时间O(b^(h+1)-1) 空间 O(h) 递归栈空间 对于二叉树b=2