LeetCode 103: Binary Tree Zigzag Level Order Traversal

Difficulty: 4

Frequency: 3

Problem:

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]]

Solution:

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > answer;        if (!root)            return answer;        list <TreeNode *> level_list[2];        vector<int> v_level;        list <TreeNode *>::iterator iter;        int i_current = 0, i_next = 1;        bool b_forward = true;        level_list[0].push_back(root);        while(!level_list[i_current].empty())        {            v_level.clear();            if (b_forward)            {                while(!level_list[i_current].empty())                {                    TreeNode * node = level_list[i_current].front();                    if (node->left)                        level_list[i_next].push_back(node->left);                    if (node->right)                        level_list[i_next].push_back(node->right);                    v_level.push_back(node->val);                    level_list[i_current].pop_front();                }                answer.push_back(v_level);                i_current = i_next;                i_next ^= 1;                b_forward = !b_forward;            }            else            {                while(!level_list[i_current].empty())                {                    TreeNode * node = level_list[i_current].back();                    if (node->right)                        level_list[i_next].push_front(node->right);                    if (node->left)                        level_list[i_next].push_front(node->left);                    v_level.push_back(node->val);                    level_list[i_current].pop_back();                }                answer.push_back(v_level);                i_current = i_next;                i_next ^=1;                b_forward = !b_forward;            }        }        return answer;    }};

Notes: