zoukankan      html  css  js  c++  java
  • LeetCode

    Binary Tree Level Order Traversal II

    2014.1.8 01:45

    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],
    ]

    confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

    Solution:

      This is a variation from: LeetCode - Binary Tree Level Order Traversal.

      Level-order traversal of a binary tree can be done with stl-queue. But in this problem it can be done with preorder traversal, too.

      When the traversal is done, reverse the result by row and you get what you need.

      Time and space complexities are both O(n), where n is the number of nodes in the tree. Space complexity comes from the local parameters passed in function calls.

    Accepted code:

     1 // 1CE, 1AC, when copying the code, make sure you don't mix up the name of function and variables!!
     2 /**
     3  * Definition for binary tree
     4  * struct TreeNode {
     5  *     int val;
     6  *     TreeNode *left;
     7  *     TreeNode *right;
     8  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     9  * };
    10  */
    11 class Solution {
    12 public:
    13     //1CE here, levelOrder, levelOrderBottom, check the names before you copy the codes!!!!
    14     // That was stupid, man~
    15     vector<vector<int> > levelOrderBottom(TreeNode *root) {
    16         // IMPORTANT: Please reset any member data you declared, as
    17         // the same Solution instance will be reused for each test case.
    18         // I could use pre-order traversal to do this.
    19         // Level-order traversal makes sense too.
    20         
    21         // 1CE here, declaration of int i is MISSSING
    22         for(int i = 0; i < result.size(); ++i){
    23             result[i].clear();
    24         }
    25         result.clear();
    26         
    27         if(root == nullptr){
    28             return result;
    29         }
    30         
    31         preOrder(root, 0);
    32 
    33         vector<int> tmp;
    34         int i = 0;
    35         while(i < result.size() - 1 - i){
    36             tmp = result[i];
    37             result[i] = result[result.size() - 1 - i];
    38             result[result.size() - 1 - i] = tmp;
    39             ++i;
    40         }
    41         return result;
    42     }
    43 private:
    44     vector<vector<int>> result;
    45     
    46     void preOrder(TreeNode *root, int height) {
    47         if(root == nullptr){
    48             return;
    49         }
    50         // 1CE here,  ) MISSING!!!
    51         while(result.size() <= height){
    52             result.push_back(vector<int>());
    53         }
    54         result[height].push_back(root->val);
    55         
    56         if(root->left != nullptr){
    57             preOrder(root->left, height + 1);
    58         }
    59         if(root->right != nullptr){
    60             // 1CE here, spelling error, pre-order preOrder
    61             // Don't rely on auto-complete, you're spoiled!!!
    62             preOrder(root->right, height + 1);
    63         }
    64     }
    65 };
  • 相关阅读:
    题解+补题
    信息安全导论期末复习
    Codeforces Round #104 (Div.2)
    中国计量大学现代科技学院第四届“中竞杯”程序设计校赛(同步赛)
    第一章练习-1
    【练习】购物车程序
    【转】Python中设置输出文字的颜色
    字符串,列表,元组,字典间的互相转换
    【转】Python Enhancement Proposal #8【PEP8】
    【转】pycharm的一些快捷键
  • 原文地址:https://www.cnblogs.com/zhuli19901106/p/3509991.html
Copyright © 2011-2022 走看看