[LeetCode] 103. Binary Tree Zigzag Level Order Traversal 二叉树的之字形层序遍历

本题是 102 题的简单变形

方法一：

     将102题的结果的偶数层节点（第2，4，6，...）vector内的元素翻转下即可；数组的翻转可以直接使用c++ stl 的std::reverse实现，

代码如下：

    

 1 /*
 2  * @Descripttion: 
 3  * @version: 
 4  * @Author: wangxf
 5  * @Date: 2020-07-07 20:14:29
 6  * @LastEditors: Do not edit
 7  * @LastEditTime: 2020-07-07 22:19:22
 8  */ 
 9 /*
10  * @lc app=leetcode.cn id=103 lang=cpp
11  *
12  * [103] 二叉树的锯齿形层次遍历
13  */
14 
15 // @lc code=start
16 /**
17  * Definition for a binary tree node.
18  * struct TreeNode {
19  *     int val;
20  *     TreeNode *left;
21  *     TreeNode *right;
22  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
23  * };
24  */
25 struct NodeInfo
26 {
27     TreeNode* ptr = nullptr;
28     int level = 0;
29     NodeInfo(TreeNode* node_ptr,int node_level):
30     ptr(node_ptr),level(node_level)
31     {}
32 };
33 
34 class Solution {
35 public:
36     vector<vector<int>> zigzagLevelOrder(TreeNode* root)
37     {
38         std::vector<vector<int>> res;
39         if(!root) return res;
40         std::queue<NodeInfo> q;
41         NodeInfo root_node(root,0);
42         q.push(root_node);
43         while(!q.empty())
44         {
45             NodeInfo cur_node = q.front();
46             q.pop();
47             int cur_level = cur_node.level;
48             int cur_node_val = cur_node.ptr->val;
49             if(res.size()<cur_level+1)
50             {
51                 vector<int> tempVec;
52                 res.push_back(tempVec);
53             }
54             res[cur_level].push_back(cur_node_val);
55             if(cur_node.ptr->left)
56             {
57                NodeInfo leftNode(cur_node.ptr->left,cur_level+1);
58                q.push(leftNode);
59             }
60             if(cur_node.ptr->right)
61             {
62                NodeInfo rightNode(cur_node.ptr->right,cur_level+1);
63                q.push(rightNode);
64             }
65         }
66         int res_size = res.size();
67         for(int i = 0;i<res_size;++i)
68         {
69             if(i%2!=0){
70                 std::reverse(res[i].begin(),res[i].end());
71             }
72         }
73         return res;
74     }
75 };
76 // @lc code=end

 

posted @ 2020-07-08 00:21  谁在写西加加  阅读(121)  评论(0)  编辑  收藏  举报