LeetCode之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]]

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:/************************************************************************    vector<vector<int> > levelOrder(TreeNode *root) {          vector<vector<int>> rst; queue<TreeNode*> currentLevel, nextLevel; currentLevel.push(root); vector<int> ivec;  if (!root) return rst; while (!currentLevel.empty()) { TreeNode *currNode = currentLevel.front(); currentLevel.pop(); if (currNode) { ivec.push_back(currNode->val); if(currNode->left)    nextLevel.push(currNode->left); if(currNode->right)   nextLevel.push(currNode->right); } if (currentLevel.empty()) { rst.push_back(ivec); ivec.clear(); swap(currentLevel, nextLevel); } } return rst;    }**************************************************************************/    vector<vector<int> > levelOrder(TreeNode *root) {        vector<vector<int>> rst;        DFS(root,0,rst);        return rst;    }        void DFS(TreeNode *root, int level, vector<vector<int> > &rst) {        if(!root)return ;        if(level>=rst.size()) rst.push_back(vector<int>());        rst[level].push_back(root->val);        DFS(root->left,level+1,rst);        DFS(root->right,level+1,rst);    }};