[LeetCode] 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:
    queue<TreeNode*> q; //存放当前层的结点
    
    vector<vector<int> >res;//存放最终的结果

    vector<vector<int> > levelOrder(TreeNode *root) {
        
        if(root==NULL)
            return this->res;
        TreeNode *p = root;
        q.push(p);
        
        fun();
        return res;
    }
private:
    void fun()
    {
        if(this->q.empty())
            return;
        TreeNode *pte;
        queue<TreeNode*> qnext;//存放下一层的结点
        vector<int> te;
        while(!this->q.empty() )
        {
               pte = this->q.front();
               this->q.pop();
               te.push_back(pte->val);
               if(pte->left!=NULL)
               {
                   qnext.push(pte->left);
               }
               if(pte->right != NULL)
               {
                   qnext.push(pte->right);
               }
        }
        res.push_back(te);
        this->q = qnext;
        fun();    
    }
};

思路：用queue存放本层所有的结点，把本层处理完清空queue再存放下一层所有的结点。