Leetcode: 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]
]

用stack, 从左到右，从右到左，轮着先入栈就行。

/**
 * 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> > result;
        if (root == NULL) {
            return result;
        }
        
        bool from_left = true;
        TreeNode *cur;
        stack<TreeNode*> nodes, tmp;
        nodes.push(root);
        
        do {
            vector<int> level;
            while (!nodes.empty()) {
                cur = nodes.top();
                nodes.pop();
                level.push_back(cur->val);
                if (from_left) {
                    if (cur->left != NULL) {
                        tmp.push(cur->left);
                    }
                    if (cur->right != NULL) {
                        tmp.push(cur->right);
                    }
                }
                else {
                    if (cur->right != NULL) {
                        tmp.push(cur->right);
                    }
                    if (cur->left != NULL) {
                        tmp.push(cur->left);
                    }
                }
            }
            result.push_back(level);
            swap(nodes, tmp);
            from_left = !from_left;
         } while (!nodes.empty());
         
         return result;
    }
};

=====================第二次========================

/**
 * 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>> result;
        bool from_left = true;
        stack<TreeNode*> first;
        stack<TreeNode*> second;
        first.push(root);
        
        vector<int> level;
        TreeNode *cur = NULL;
        while (!first.empty()) {
            while (!first.empty()) {
                cur = first.top();
                first.pop();
                if (cur != NULL) {
                    level.push_back(cur->val);
                    if (from_left) {
                        second.push(cur->left);
                        second.push(cur->right);
                    }
                    else {
                        second.push(cur->right);
                        second.push(cur->left);
                    }
                }
            }
            
            if (!level.empty()) {
                result.push_back(level);
                level.clear();
            }
            first.swap(second);
            from_left = !from_left;
        }
        
        return result;
    }
};