AC Log - bfs
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link: https://aclog.neix.in/index.php/tag/bfs/
items: 二叉树的右视图

二叉树的右视图
type: item
date: 2025-12-18T05:30:56+00:00
description: 题目链接：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。方法一：层序遍历/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: vector<int> rightSideView(TreeNode* root) { vector<int> ans; if (root == nullptr) return ans; queue<TreeNode*> q; q.push(root); int num = 0; while (q.size()) { int t = q.size(); //int num; while (t--) { TreeNode* tmp = q.front(); q.pop(); num = tmp->val; if (tmp->left != nullptr) q.push(tmp->left); if (tmp->right != nullptr) q.push(tmp->right); } ans.push_back(num); } return ans; } };方法二：bfs/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: vector<int> rightSideView(TreeNode* root) { vector<int> ans; if (!root) return ans; queue<TreeNode*> q; q.push(root); while (q.size()) { int size = q.size(); ans.push_back(q.front()->val); for (int i = 0;i < size;i++) { TreeNode* node = q.front(); q.pop(); if (node->right) q.push(node->right); if (node->left) q.push(node->left); } } return ans; } };方法三：dfs/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: void dfs(TreeNode* root,int depth,vector<int>& ans) { if (root == nullptr) return ; if (depth == ans.size()) ans.push_back(root->…