【Binary Tree Level Order Traversal II 】cpp

题目:

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

代码:

复制代码
/**
 * 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<vector<int>> levelOrderBottom(TreeNode* root) {
            vector<vector<int> > ret;
            if (!root) return ret;
            vector<int> tmp_ret;
            deque<TreeNode *> currLevel, nextLevel;
            currLevel.push_back(root);
            while ( !currLevel.empty() )
            {
                while ( !currLevel.empty() )
                {
                    TreeNode * tmp = currLevel.front();
                    currLevel.pop_front();
                    tmp_ret.push_back(tmp->val);
                    if ( tmp->left ) nextLevel.push_back(tmp->left);
                    if ( tmp->right ) nextLevel.push_back(tmp->right);
                }
                ret.push_back(tmp_ret);
                tmp_ret.clear();
                std::swap(currLevel, nextLevel);
            }
            std::reverse(ret.begin(), ret.end());
            return ret;
    }
};
复制代码

tips:

Binary Tree Level Order Traversal的基础上加一个reverse即可。

============================================

第二次过这道题,直接用reverse的路子了。

复制代码
/**
 * 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<vector<int>> levelOrderBottom(TreeNode* root) {
            vector<vector<int> > ret;
            queue<TreeNode*> curr;
            queue<TreeNode*> next;
            if ( root ) curr.push(root);
            while ( !curr.empty() )
            {
                vector<int> tmp;
                while ( !curr.empty() )
                {
                    tmp.push_back(curr.front()->val);
                    if ( curr.front()->left ) next.push(curr.front()->left);
                    if ( curr.front()->right ) next.push(curr.front()->right);
                    curr.pop();
                }
                ret.push_back(tmp);
                std::swap(next, curr);
            }
            std::reverse(ret.begin(), ret.end());
            return ret;
    }
};
复制代码

 

posted on   承续缘  阅读(143)  评论(0编辑  收藏  举报

编辑推荐:
· go语言实现终端里的倒计时
· 如何编写易于单元测试的代码
· 10年+ .NET Coder 心语,封装的思维:从隐藏、稳定开始理解其本质意义
· .NET Core 中如何实现缓存的预热?
· 从 HTTP 原因短语缺失研究 HTTP/2 和 HTTP/3 的设计差异
阅读排行:
· 周边上新:园子的第一款马克杯温暖上架
· Open-Sora 2.0 重磅开源!
· 分享 3 个 .NET 开源的文件压缩处理库,助力快速实现文件压缩解压功能!
· Ollama——大语言模型本地部署的极速利器
· DeepSeek如何颠覆传统软件测试?测试工程师会被淘汰吗?

导航

< 2025年3月 >
23 24 25 26 27 28 1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 28 29
30 31 1 2 3 4 5

统计

点击右上角即可分享
微信分享提示