199. Binary Tree Right Side View

最新推荐文章于 2025-01-21 20:19:54 发布

YHJunE

最新推荐文章于 2025-01-21 20:19:54 发布

阅读量185

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode刷题

本文链接：https://blog.youkuaiyun.com/yanhangy/article/details/78073174

leetcode刷题专栏收录该内容

33 篇文章

订阅专栏

本文介绍了一种算法，用于获取二叉树从顶部到底部的右侧节点值。通过两种方法实现：广度优先搜索（BFS）和深度优先搜索（DFS）。这两种方法都能有效地遍历树并返回所需的节点值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

For example:
Given the following binary tree,

   1            <---
 /   \
2     3         <---
 \     \
  5     4       <---

You should return [1, 3, 4].

Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int> res;
        if(!root){
            return res;
        }
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            int size = q.size();
            for(int i = 0; i < size; ++i){
                TreeNode* temp = q.front();
                q.pop();
                
                if(i == size - 1){
                    res.push_back(temp->val);
                }
                
                if(temp->left) q.push(temp->left);
                if(temp->right) q.push(temp->right);
            }
            
        }
        return res;
    }
};

class Solution {
public:
    void recursion(TreeNode *root, int level, vector<int> &res)
    {
        if(root==NULL) return ;
        if(res.size()<level) res.push_back(root->val);
        recursion(root->right, level+1, res);
        recursion(root->left, level+1, res);
    }
    
    vector<int> rightSideView(TreeNode *root) {
        vector<int> res;
        recursion(root, 1, res);
        return res;
    }
};