103. Binary Tree Zigzag Level Order Traversal

问题：

求二叉树的正反交替层序遍历。

第一层从左向右，第二次从右向左...

Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[3],[20,9],[15,7]]

Example 2:
Input: root = [1]
Output: [[1]]

Example 3:
Input: root = []
Output: []
 
Constraints:
The number of nodes in the tree is in the range [0, 2000].
-100 <= Node.val <= 100

　　

Example 1:

解法：BFS

queue：保存每一层遍历节点。

for为一层。

flg记录奇偶层，奇数层再使用reverse，将一层顺序反转。

代码参考：

/**
 * 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) {
        queue<TreeNode*> q;
        vector<vector<int>> res;
        int flg = 0;
        if(root) q.push(root);
        while(!q.empty()) {
            int sz = q.size();
            vector<int> curlevel;
            for(int i=0; i<sz; i++) {
                TreeNode* cur = q.front();
                q.pop();
                curlevel.push_back(cur->val);
                if(cur->left) q.push(cur->left);
                if(cur->right) q.push(cur->right);
            }
            if(!curlevel.empty() && flg%2) {
                reverse(curlevel.begin(), curlevel.end());
            }
            if(!curlevel.empty()) res.push_back(curlevel);
            flg++;
        }
        return res;
    }
};