Binary Tree Level Order Traversal

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

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

    3
   / 
  9  20
    /  
   15   7

return its level order traversal as:

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

解题思路:能够用DFS和BFS思想来解这题.DFS仅仅须要递归遍历深度，将相应深度的值按顺序输出.在用BFS分层遍历时候。一開始内存超了。后来增加深度信息AC了

#include<iostream>
#include<vector>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;

struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
//内存过大了
vector<vector<int> > levelOrder(TreeNode *root) {
	queue<TreeNode*>  BFS_Queue;
	vector<int> LevelVal;
	vector<TreeNode*> LevelNode;
	vector<vector<int> > Result;

	if (root == NULL)
		return Result;
	BFS_Queue.push(root);
	
	while (!BFS_Queue.empty())
	{
		while (!BFS_Queue.empty())
		{
			TreeNode* Curnode = BFS_Queue.front();
			LevelVal.push_back(Curnode->val);
			BFS_Queue.pop();
			if (Curnode->left)
				LevelNode.push_back(Curnode->left);
			if (Curnode->right)
				LevelNode.push_back(Curnode->right);
		}
		Result.push_back(LevelVal);
		for_each(LevelNode.begin(), LevelNode.end(), [&BFS_Queue](TreeNode* a){BFS_Queue.push(a); });
		LevelVal.clear();
	}
	return Result;
	
}
//引入深度信息：AC
vector<vector<int> > levelOrder(TreeNode *root) {
	queue<pair<TreeNode*,int> >  BFS_Queue;
	vector<int> LevelVal;
	vector<vector<int> > Result;

	if (root == NULL)
		return Result;
	BFS_Queue.push(make_pair(root,0));
	int Curlevel = 0;
	while (!BFS_Queue.empty())
	{
		TreeNode* Curnode = BFS_Queue.front().first;
		if (BFS_Queue.front().second != Curlevel)
		{
			Result.push_back(LevelVal);
			LevelVal.clear();
			Curlevel = BFS_Queue.front().second;
		}
		LevelVal.push_back(Curnode->val);
		BFS_Queue.pop();
		if (Curnode->left)
			BFS_Queue.push(make_pair(Curnode->left, Curlevel + 1));
		if (Curnode->right)
			BFS_Queue.push(make_pair(Curnode->right, Curlevel + 1));
	}
	Result.push_back(LevelVal);
	return Result;
}