Print a binary tree in an m*n 2D string array following these rules:

  1. The row number m should be equal to the height of the given binary tree.
  2. The column number n should always be an odd number.
  3. The root node's value (in string format) should be put in the exactly middle of the first row it can be put. The column and the row where the root node belongs will separate the rest space into two parts (left-bottom part and right-bottom part). You should print the left subtree in the left-bottom part and print the right subtree in the right-bottom part. The left-bottom part and the right-bottom part should have the same size. Even if one subtree is none while the other is not, you don't need to print anything for the none subtree but still need to leave the space as large as that for the other subtree. However, if two subtrees are none, then you don't need to leave space for both of them.
  4. Each unused space should contain an empty string "".
  5. Print the subtrees following the same rules.

Example 1:

Input:
1
/
2
Output:
[["", "1", ""],
["2", "", ""]]

Example 2:

Input:
1
/ \
2 3
\
4
Output:
[["", "", "", "1", "", "", ""],
["", "2", "", "", "", "3", ""],
["", "", "4", "", "", "", ""]]

Example 3:

Input:
1
/ \
2 5
/
3
/
4
Output: [["", "", "", "", "", "", "", "1", "", "", "", "", "", "", ""]
["", "", "", "2", "", "", "", "", "", "", "", "5", "", "", ""]
["", "3", "", "", "", "", "", "", "", "", "", "", "", "", ""]
["4", "", "", "", "", "", "", "", "", "", "", "", "", "", ""]]

Note: The height of binary tree is in the range of [1, 10].

思路:

观察例子可以发现,行m就是树的高度h,列n就是2h-1,那么首先构造一个二维数组ret[m][n].用“”初始化。

观察得到,每一个深度为d的树节点t都在ret所在的行为d(假设下标从1开始)的中间。而t的左孩子,则深度为t+1,所在的行下标范围[left,mid-1]

可以递归。

int maxDepth(TreeNode *root)
{
int depth;
if(root==NULL)return ;
else{
depth=(maxDepth(root->left)>maxDepth(root->right)?
maxDepth(root->left):maxDepth(root->right))+;
return depth;
}
}
void printTree(vector<vector<string>>& ret,TreeNode* root,int leftIndex,int rightIndex,int curDepth,int maxDepth)
{
if(curDepth>maxDepth || root==NULL||leftIndex>rightIndex) return;
int mid = (leftIndex+rightIndex)/;
ret[curDepth-][mid] = to_string(root->val);
printTree(ret,root->left,leftIndex,mid-,curDepth+,maxDepth);
printTree(ret,root->right,mid+,rightIndex,curDepth+,maxDepth);
}
vector<vector<string>> printTree(TreeNode* root)
{
int depth = maxDepth(root);
if(depth == ) return {{}}; vector<vector<string>> ret(depth,vector<string>(pow(,depth)-,"")); printTree(ret,root,,pow(,depth)--,,depth);
return ret;
}

[leetcode-655-Print Binary Tree]的更多相关文章

  1. [LeetCode] 655. Print Binary Tree 打印二叉树

    Print a binary tree in an m*n 2D string array following these rules: The row number m should be equa ...

  2. LeetCode 655. Print Binary Tree (C++)

    题目: Print a binary tree in an m*n 2D string array following these rules: The row number m should be ...

  3. LC 655. Print Binary Tree

    Print a binary tree in an m*n 2D string array following these rules: The row number m should be equa ...

  4. 【LeetCode】655. Print Binary Tree 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 DFS BFS 日期 题目地址:https://le ...

  5. [LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树

    Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that ...

  6. 【LEETCODE OJ】Binary Tree Postorder Traversal

    Problem Link: http://oj.leetcode.com/problems/binary-tree-postorder-traversal/ The post-order-traver ...

  7. 【一天一道LeetCode】#107. Binary Tree Level Order Traversal II

    一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源: htt ...

  8. 【一天一道LeetCode】#103. Binary Tree Zigzag Level Order Traversal

    一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源: htt ...

  9. C++版 - 剑指offer 面试题39:判断平衡二叉树(LeetCode 110. Balanced Binary Tree) 题解

    剑指offer 面试题39:判断平衡二叉树 提交网址:  http://www.nowcoder.com/practice/8b3b95850edb4115918ecebdf1b4d222?tpId= ...

  10. (二叉树 递归) leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

    Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that ...

随机推荐

  1. checkbox 全选

    <template> <div class="hello"> <table> <tr> <th><input ty ...

  2. JavaWeb遗漏的知识点

    1. String javax.servlet.ServletContext.getRealPath(String path)方法 官方文档解释:Gets the real path correspo ...

  3. iOS之苹果调整 App Store 截图上传规则,截图尺寸、大小等

    作者:ASO100链接:https://zhuanlan.zhihu.com/p/23041522来源:知乎著作权归作者所有.商业转载请联系作者获得授权,非商业转载请注明出处. 自从 8 月中旬苹果向 ...

  4. php第一节(入门语法、数据类型)

    <?php /** * 变量命名用 $ 符 * 变量名称的命名规范 * 1.变量名称以$标示 * 2.变量名称只能以字母和下划线开头 * 3.变量的名称只能包含字母.下划线.数字 * 4.变量名 ...

  5. DISTINCT 去重仍有重复的分析

    logger日志报错 插入数据时违反主键唯一约束 org.springframework.dao.DuplicateKeyException: ### Error updating database. ...

  6. MySQL的日志相关内容

    本篇文章介绍一下mysql的备份和日志,由于备份时需要用到日志,所以在讲备份前,如果日志内容篇幅过长,将会把日志和备份分开单独来讲,先简单介绍一下mysql的日志相关内容. MySQL日志 日志是my ...

  7. 简单使用PuTTy登录centos虚拟机

    博主刚刚开始学习Linux,想通过写博客的方式加深对linux学习的记忆 使用虚拟机安装精简版的linux后发现窗口字体太小,于是就想着通过PuTTy登录的方式解决 简单安装linux后 将linux ...

  8. yum仓库客户端搭建和NTP时间同步客户端配置

    一.yum仓库客户端搭建 yum源仓库搭建分为服务器端和客户端. 服务端主要提供软件(rpm包)和yumlist.也就是提供yum源的位置.一般是通过http或者ftp提供位置. 客户端的配置:yum ...

  9. ELK 安装部署实战 (最新6.4.0版本)

    一.实战背景 根据公司平台的发展速度,对于ELK日志分析日益迫切.主要的需求有: 1.用户行为分析 2.运营活动点击率分析 作为上述2点需求,安装最新版本6.4.0是非常有必要的,大家可根据本人之前博 ...

  10. Go web表单验证

    开发Web的一个原则就是,不能信任用户输入的任何信息,所以验证和过滤用户的输入信息就变得非常重要 必填字段 if len(r.Form["username"][0])==0{ // ...