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,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its zigzag level order traversal as: [ [3], [20,9], [15,7] ] confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * Definition for binary tree
 * 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) {
        vector<vector<int> > res;
        if (root == NULL) return res;
        res.push_back(vector<int>(1, root->val));
        
        vector<TreeNode*> pre_level(1, root);
        vector<TreeNode*> cur_level;

        bool l2r = true;
        
        while (!pre_level.empty()) {
            res.push_back(vector<int>());
            int plen = pre_level.size();
            
            TreeNode* node = NULL;
            
            for (int i=0; i<plen; i++) {
                node = pre_level[i];
                if (node->left != NULL) {
                    res.back().push_back(node->left->val);
                    cur_level.push_back(node->left);
                }
                if (node->right != NULL) {
                    res.back().push_back(node->right->val);
                    cur_level.push_back(node->right);
                }
            }
            if (l2r) {
                reverse(res.back().begin(), res.back().end());
                //reverse(cur_level.begin(), cur_level.end());
            }
            
            pre_level.clear();
            swap(pre_level, cur_level);
            
            if (pre_level.empty()) res.pop_back();
            l2r = !l2r;
        }
        
        return res;
    }
};

esay

第二轮：

/**
 * 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>> zigzagLevelOrder(TreeNode* root) {
        vector<vector<int> > res;
        
        if (root == NULL) {
            return res;
        }
        
        vector<TreeNode*> last;
        vector<TreeNode*> next;
        
        res.push_back(vector<int>(1, root->val));
        last.push_back(root);
        vector<int> curr;
        int level = 1;
        
        while (!last.empty()) {
            next.clear();
            curr.clear();
            for(TreeNode* node : last) {
                if (node->left != NULL) {
                    next.push_back(node->left);
                    curr.push_back(node->left->val);
                }
                if (node->right != NULL) {
                    next.push_back(node->right);
                    curr.push_back(node->right->val);
                }
            }
            if (curr.empty()) {
                break;
            }
            if (level++ & 0x1) {
                reverse(curr.begin(), curr.end());
            }
            swap(next, last);
            res.push_back(curr);
        }
        return res;
    }
};