LeetCode-103. 二叉树的锯齿形层次遍历

给定一个二叉树，返回其节点值的锯齿形层次遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。

例如：
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7
返回锯齿形层次遍历如下：

[
  [3],
  [20,9],
  [15,7]
]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-zigzag-level-order-traversal
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在102题目的基础上，根据奇偶顺序进行反转。

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

/*

        3
       / \
      9  20
     / \
    15  7

   [
    [3],
    [20,9],
    [15,7]
   ]

*/

/*
 *[1,2,3,4,null,null,5]
 *     1
 *    / \
 *   2   3
 *  / \
 * 4   5
 */

/*

[1,2,3,4,null,null,5]

[[1],[3,2],[5,4]]

[[1],[3,2],[4,5]]

*/

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        if(root==NULL)
            return vec_res;

        tree_queue.push(root);
        deepcount = 1;

        while(!tree_queue.empty()){
            int size = tree_queue.size();

            vec_temp.clear();

            while(size--){
                TreeNode* front = tree_queue.front();
                tree_queue.pop();

                if(front->left){
                    tree_queue.push(front->left);
                }
                if(front->right){
                    tree_queue.push(front->right);
                }
                vec_temp.push_back(front->val);
            }

            /* reverse changed by tree level */
            if((deepcount++)%2==0){
                reverse(vec_temp.begin(),vec_temp.end());
            }

            vec_res.push_back(vec_temp);
        }
        return vec_res;
    }
private:
    vector<int> vec_temp;
    vector<vector<int>> vec_res;
    queue<TreeNode*> tree_queue;
    int deepcount;

};

int main(){
    TreeNode *root = new TreeNode(1);
    TreeNode *l1 = new TreeNode(2);
    TreeNode *r1 = new TreeNode(3);
    TreeNode *rl2 = new TreeNode(4);
    TreeNode *rr2 = new TreeNode(5);

    root->left = l1;
    root->right = r1;
    l1->left = rl2;
    l1->right = rr2;

    Solution *ps = new Solution();
    vector<vector<int>> res = ps->zigzagLevelOrder(root);
    for(int i=0;i<res.size();i++){
        for(int j=0;j<res[i].size();j++){
            printf("%d ",res[i][j]);
        }
        printf("\n");
    }

    return 0;
}