LeetCode | Binary Tree Zigzag Level Order Traversal

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二叉树Z型输出

Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).

For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]

其实还是广度优先遍历，只不过这里对偶数行需要reverse一下，
发现stl可以方便地完成reverse(row.begin(),row.end())即可

然后也要注意这类题目需要知道递归的怎么写。
其实递归就是传入一个level来判断当前进入到了第几层，然后再将数据存入到这一层中

class Solution {
public://递归版本vector<vector<int>> zigzagLevelOrder(TreeNode* root) {vector<vector<int> > result;tranverse(root,0,result);for(int i=0;i<result.size();i++)if(i&0x01) reverse(result[i].begin(),result[i].end());return result;}void tranverse(TreeNode* root,int level,vector<vector<int> > &result){if(root==NULL) return;if(level>=result.size())result.push_back(vector<int> ());result[level].push_back(root->val);tranverse(root->left,level+1,result);tranverse(root->right,level+1,result);}//非递归版
//     vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
//         vector<vector<int> > result; 
//      if(!root) return result;
//      queue<TreeNode*> Q;
//      Q.push(root);//      int count=0;
//      while(!Q.empty()){
//          queue<TreeNode*> Qnext;
//          vector<int> row;
//          while(!Q.empty()){
//              TreeNode* node=Q.front();
//              Q.pop();//              row.push_back(node->val);//              if(node->left) Qnext.push(node->left);
//              if(node->right) Qnext.push(node->right);
//          }
//          if(count & 0x01) reverse(row.begin(),row.end());
//          result.push_back(row);
//          Q=Qnext;
//          count++;
//      }
//      return result;
//     }
};

字符串Z型输出——原题

上面的题是由这道题演化而来的
The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P A H N
A P L S I I G
Y I R
And then read line by line: “PAHNAPLSIIGYIR”
Write the code that will take a string and make this conversion given a number of rows:

string convert(string text, int nRows);
convert(“PAYPALISHIRING”, 3) should return “PAHNAPLSIIGYIR”.

题意很简单，很像之前让自己排列并输出一些数据

于是主要就是找规律了~

class Solution {
public:string convert(string s, int numRows) {int n=s.size();if(n<=1) return s;if(numRows<=1) return s;string target="";for(int i=0;i<numRows;i++){for(int j=i,count=1;j<n;j+=2*numRows-2,count++){target=target+s[j];if(j+(numRows-i-1)*2<n && i>0 && i>0 && i<numRows-1)target=target+s[j+(numRows-i-1)*2];}}return target;}
};