学步园

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]
]

思路：

通过迭代方法按行访问。然后每访问一行，改变一次访问顺序和存放子结点的顺序。

题解：

/**
 * 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>> ret;
        
        if (root == nullptr)
            return ret;
        
        // direction of visiting
        bool fwd_direction = false;
        
        list<TreeNode*> this_row;
        list<TreeNode*> next_row;
        
        this_row.push_back(root);
        ret.push_back(vector<int>());
        
        while(!this_row.empty())
        {
            TreeNode* l;
            
            if (fwd_direction) 
            {
                l = this_row.front();
                this_row.pop_front();
            }
            else
            {
                l = this_row.back();
                this_row.pop_back();
            }
            
            ret.back().push_back(l->val);
            
            if (fwd_direction)
            {
                if (l->right != nullptr) next_row.push_back(l->right);
                if (l->left != nullptr) next_row.push_back(l->left);
            }
            else
            {
                if (l->left != nullptr) next_row.push_front(l->left);
                if (l->right != nullptr) next_row.push_front(l->right);
            }
            
            if (this_row.empty())
            {
                swap(this_row, next_row);
                ret.push_back(vector<int>());
                fwd_direction = !fwd_direction;
            }
        }
        
        //if (ret.back().empty()) 
        ret.pop_back();
        
        return ret;
    }
};