LeetCode 107: Binary Tree Level Order Traversal II

Difficulty: 3

Frequency: 1

Problem:

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

[  [15,7]  [9,20],  [3],]

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> > levelOrderBottom(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > answer;        if (root==NULL)            return answer;                    list <vector<int> > stack;           vector <int> one_level;        vector<TreeNode *> level[2];        int i_level_num = 0;        level[i_level_num].push_back(root);        while (!level[i_level_num].empty())        {            int i_next_level = i_level_num^1;            for (int i = 0; i<level[i_level_num].size(); i++)            {                if (level[i_level_num][i]->left!=NULL)                    level[i_next_level].push_back(level[i_level_num][i]->left);                                    if (level[i_level_num][i]->right!=NULL)                    level[i_next_level].push_back(level[i_level_num][i]->right);                                    one_level.push_back(level[i_level_num][i]->val);            }            level[i_level_num].clear();            stack.push_back(one_level);            one_level.clear();            i_level_num = i_next_level;        }        while(!stack.empty())        {            answer.push_back(stack.back());            stack.pop_back();        }        return answer;    }};

Notes:

Use a stack to store and reverse data.