leetcode_Binary Tree Level Order Traversal _easy

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]
]

方法1：先根节点，在左右子树进行递归，然后合并三者的结果---用时较长，方法不够好

/**
 * 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> > ivv;
        if(root!=NULL)
        {
            vector<vector<int> > ivvL,ivvR;
            ivvL=levelOrder(root->left);//递归，得到左子树的结果
            ivvR=levelOrder(root->right);//递归,得到右子树的结果
            
            //合并根、左右子树的结果到ivv
            vector<int> ivROOT;
            ivROOT.push_back(root->val);
            ivv.push_back(ivROOT);//现将根元素放入
            vector<vector<int> >::iterator iter1,iter2,iter3;
            for( iter1=ivvL.begin(),iter2=ivvR.begin(); iter1!=ivvL.end() && iter2!=ivvR.end(); iter1++,iter2++)
            {
                (*iter1).insert((*iter1).end(),(*iter2).begin(),(*iter2).end());//合并到ivvL中的vector<int>对应元素中去
                ivv.push_back(*iter1);       
            }
            iter3=iter1==ivvL.end()?iter2:iter1;
            for(; (iter3!=ivvL.end())&&(iter3!=ivvR.end()); iter3++)
                ivv.push_back(*iter3);//将剩下的元素都加入进去
        }
        return ivv;
    }
};
方法2：参考了戴同学的方法，直接利用层数作为参数来传递是关键，速度快多了。

/**
 * 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> > ivv;
        traverse(root,0,ivv);
        return ivv;
    }
    
    void traverse(TreeNode *root,int depth,vector<vector<int> > &ivv)
    {
        if(root!=NULL)
        {
            if(ivv.size()<depth+1)
            {
                vector<int> iv;
                ivv.push_back(iv);
            }
            ivv[depth].push_back(root->val);
            traverse(root->left,depth+1,ivv);
            traverse(root->right,depth+1,ivv);
        }
    }
};