Binary Tree Level Order Traversal II.cpp

/*
    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).

    Link: http://www.lintcode.com/en/problem/binary-tree-level-order-traversal-ii/

    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: Use reverse the result of part one problem.

    Source: https://github.com/kamyu104/LintCode/blob/master/C%2B%2B/binary-tree-level-order-traversal-ii.cpp
*/

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 
 
class Solution {
    /**
     * @param root : The root of binary tree.
     * @return : buttom-up level order a list of lists of integer
     */
public:
    vector<vector<int>> levelOrderBottom(TreeNode *root) {
        // write your code here
        vector<vector<int>> output;
        queue<TreeNode *> que;
        vector<int> cur_level;

        // Enqueue when node is not nullptr.
        if (root) {
            que.emplace(root);
        }

        int cur_level_cnt = que.size();
        while (!que.empty()) {
            TreeNode *node = que.front();
            que.pop();
            --cur_level_cnt;

            cur_level.emplace_back(node->val);

            // Enqueue the next level.
            if (node->left) {
                que.emplace(node->left);
            }
            if (node->right) {
                que.emplace(node->right);
            }

            // Current level has been all visited.
            if (cur_level_cnt == 0) {
                cur_level_cnt = que.size();
                output.emplace_back(move(cur_level));
            }
        }
        reverse(output.begin(), output.end());
        return output;
    }
};