[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：

    

/*
 * @Descripttion: 
 * @version: 
 * @Author: wangxf
 * @Date: 2020-07-07 20:14:29
 * @LastEditors: Do not edit
 * @LastEditTime: 2020-07-07 22:19:22
 */ 
/*
 * @lc app=leetcode.cn id=103 lang=cpp
 *
 * [103] 二叉树的锯齿形层次遍历
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
struct NodeInfo
{
    TreeNode* ptr = nullptr;
    int level = 0;
    NodeInfo(TreeNode* node_ptr,int node_level):
    ptr(node_ptr),level(node_level)
    {}
};

class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root)
    {
        std::vector<vector<int>> res;
        if(!root) return res;
        std::queue<NodeInfo> q;
        NodeInfo root_node(root,0);
        q.push(root_node);
        while(!q.empty())
        {
            NodeInfo cur_node = q.front();
            q.pop();
            int cur_level = cur_node.level;
            int cur_node_val = cur_node.ptr->val;
            if(res.size()<cur_level+1)
            {
                vector<int> tempVec;
                res.push_back(tempVec);
            }
            res[cur_level].push_back(cur_node_val);
            if(cur_node.ptr->left)
            {
               NodeInfo leftNode(cur_node.ptr->left,cur_level+1);
               q.push(leftNode);
            }
            if(cur_node.ptr->right)
            {
               NodeInfo rightNode(cur_node.ptr->right,cur_level+1);
               q.push(rightNode);
            }
        }
        int res_size = res.size();
        for(int i = 0;i<res_size;++i)
        {
            if(i%2!=0){
                std::reverse(res[i].begin(),res[i].end());
            }
        }
        return res;
    }
};
// @lc code=end