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]
]

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

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / 
 2   3
    /
   4
    
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

思路：这道题是二叉树的层次遍历算法，简单来说就是借用队列来实现二叉树的层次遍历，使用循环方法。每次将一层的结点放入队列中，将一层的结点遍历其左右结点，然后pop这一层的结点。

/**
 * 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> > result;
        queue<TreeNode *> pTree;
        if(root==NULL)
            return result;
        pTree.push(root);
        int count=1;
        vector<int> temp;
        while(!pTree.empty())
        {
            temp.clear();
            int temp_count=0;
            for(int i=0;i<count;i++)
            {
                TreeNode *tn=pTree.front();
                pTree.pop();
                temp.push_back(tn->val);
                if(tn->left!=NULL)
                {
                    pTree.push(tn->left);
                    temp_count++;
                }
                if(tn->right!=NULL)
                {
                    pTree.push(tn->right);
                    temp_count++;
                }
            }
            count=temp_count;
            result.push_back(temp);
        }
        return result;
    }
};

看了其他人的思路，使用递归方法实现的：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    vector<vector<int> > result;    
public:
    void traversal(TreeNode *root,int depth)
    {
        if(root==NULL)
            return;
        if(result.size()>depth)
        {
            result[depth].push_back(root->val);
        }
        else
        {
            vector<int> data;
            data.push_back(root->val);
            result.push_back(data);
        }
        traversal(root->left,depth+1);
        traversal(root->right,depth+1);
    }
    vector<vector<int> > levelOrder(TreeNode *root) {
        
        result.clear();
        traversal(root,0);
        return result;
    }
};