Binary Tree Level Order TraversalI.txt

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

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solution:
////////////////////////////////////////////
struct TreeNode {
  int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


vector<vector<int> > levelOrder(TreeNode *root)
{
	vector<vector<int> > ret;
	if(root == NULL)
		return ret;
	queue<TreeNode*> q;
	q.push(root);
	vector<int> tmp(0);	
	int count = 1;
	int level = 0;
	while(!q.empty())
	{
		tmp.clear();
		level = 0;
		for(int i = 0; i < count; ++i)
		{
			root = q.front();
			q.pop();
			tmp.push_back(root->val);
			if(root->left != NULL)
			{
				q.push(root->left);
				level++;
			}
			if(root->right != NULL)
			{
				q.push(root->right);
				level++;
			}
		}
		ret.push_back(tmp);
		count = level;
	}
	reverse(ret.begin(), ret.end());
	return ret;
}