#Leetcode# 199. Binary Tree Right Side View

二叉树右视图的层序遍历

最新推荐文章于 2025-09-11 18:44:42 发布

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最新推荐文章于 2025-09-11 18:44:42 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zlrrrr/p/10133014.html

本文详细介绍了如何使用层序遍历算法解决LeetCode上的二叉树右视图问题，通过两个不同的实现方式展示了如何从上到下返回可以看到的节点值。第一种方法使用队列进行广度优先搜索，第二种方法采用递归方式实现。两种方法均能有效获取二叉树每层最右侧节点的值。

https://leetcode.com/problems/binary-tree-right-side-view/

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       <---

代码：

/**
 * 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<vector<int> > levelOrder;
        vector<int> ans;
        if(!root) return ans;
        
        queue<pair<TreeNode*, int>> q;
        q.push(make_pair(root, 1));
        while(!q.empty()) {
            pair<TreeNode*, int> tp = q.front();
            q.pop();
            
            if(tp.second > levelOrder.size()) {
                vector<int> v;
                levelOrder.push_back(v);
            }
            
            levelOrder[tp.second - 1].push_back(tp.first->val);
            
            if(tp.first->left) 
                q.push(make_pair(tp.first->left, tp.second + 1));
            
            if(tp.first->right) 
                q.push(make_pair(tp.first->right, tp.second + 1));
        }
        
        
        for(int i = 0; i < levelOrder.size(); i ++) {
            ans.push_back(levelOrder[i][levelOrder[i].size() - 1]);
        }
        
        return ans;
    }
    
};

　　FH 写的 bfs

代码：

/**
 * 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<vector<int> > levelOrder;
        vector<int> ans;
        if(!root) return ans;
        
        helper(root, levelOrder, 1);
        
        for(int i = 0; i < levelOrder.size(); i ++) {
            ans.push_back(levelOrder[i][levelOrder[i].size() - 1]);
        }
        
        return ans;
    }
    void helper(TreeNode* root, vector<vector<int> >& ans, int depth) {
      vector<int> v;
      if(depth > ans.size()) {
        ans.push_back(v);
        v.clear();
      }
      ans[depth - 1].push_back(root -> val);
      if(root -> left)
        helper(root -> left, ans, depth + 1);
      if(root -> right)
        helper(root -> right, ans, depth + 1);
    }
};

　　小张写的 code 感觉一晚上都在看层序遍历

转载于:https://www.cnblogs.com/zlrrrr/p/10133014.html