LeetCode 103. Binary Tree Zigzag Level Order Traversal

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]
]

解析：层次遍历二叉树，之字形输出结果。还是借助队列进行层次遍历，同时使用变量来记录层数，将每层的数据放进vector中，如果是奇数层则把vector逆序。具体代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int>> result;
        if(root == NULL)
            return result;
        queue<TreeNode*> q;
        q.push(root);
        int cnt = 0;
        while(!q.empty())
        {
            int len = q.size();
            vector<int> vecs;
            for(int i=0; i < len; i++)
            {
                TreeNode* tmp = q.front();
                q.pop();
                vecs.push_back(tmp->val);
                if(tmp->left)
                    q.push(tmp->left);
                if(tmp->right)
                    q.push(tmp->right);
                
            }
            if(cnt % 2 == 1)
                std::reverse(vecs.begin(), vecs.end());
            result.push_back(vecs);
            cnt ++;
        }
        return result;
    }
};