【binary-tree-level-order-traversal-ii】

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],
]

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

思路：層次遍歷二叉樹，然後再進行逆序操作；

class Solution {
public:
	vector<vector<int>> levelOrderBottom(TreeNode* root)
	{
		vector<vector<int>> res;

		if (root == NULL)
		{
			return res;
		}

		queue<TreeNode*> q;
		q.push(root);

		while (q.size()>0)
		{
			vector<int> level;
			for (int i = 0, n = q.size(); i<n; i++)
			{
				TreeNode* tmp = q.front();
				q.pop();

				level.push_back(tmp->val);

				if (tmp->left)
				{
					q.push(tmp->left);
				}
				if (tmp->right)
				{
					q.push(tmp->right);
				}
			}

			res.push_back(level);
		}

		reverse(res.begin(), res.end());

		return res;
	}
};