一、题目

Given the root of a binary tree, return the zigzag level order traversal of its nodes’ values. (i.e., from left to right, then right to left for the next level and alternate between).

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

二、题解

/*** 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>> res;queue<TreeNode*> q;if(!root) return res;q.push(root);int level = 0;while(!q.empty()){int size = q.size();vector<int> tmp;while(size--){TreeNode* t = q.front();q.pop();tmp.push_back(t->val);if(t->left) q.push(t->left);if(t->right) q.push(t->right);}if(level % 2 == 1){reverse(tmp.begin(),tmp.end());res.push_back(tmp);}else res.push_back(tmp);level++;}return res;}
};