问题

按之字形层序遍历二叉树

算法：BFS

这道题容易想复杂，事实上，只需要考虑下之字形和正常层序遍历的区别即可：隔层反序
我们正常遍历每一层都会得到一个数组，那么只需要隔一层用reverse反转即可

时间复杂度

$O(N)$

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> q;
        vector<vector<int>> ans;
        if(!root) return ans;
        q.push(root);
        bool zigzag = false;

        while(q.size()){
            int cnt = q.size();
            vector<int> temp;
            for(int i = 0; i < cnt; i ++){
                TreeNode* u = q.front();
                q.pop();
                if(u->left)  q.push(u->left);
                if(u->right) q.push(u->right);
                temp.push_back(u->val);
            }
            if(zigzag) reverse(temp.begin(), temp.end());
            zigzag = !zigzag;
            ans.push_back(temp);
        }

        return ans;
    }
};