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;
}