Binary Tree Level Order Traversal-LeetCode(二叉树层序遍历)

Binary Tree Level Order Traversal---LeetCode(二叉树层序遍历)

题目描述：

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.

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

/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector<int> > res;        vector<int> lev;        if(root == NULL) return res;        queue<TreeNode *> que;        que.push(root);        que.push(NULL); //end of one level        while(true)        {            TreeNode *cur = que.front();            que.pop();                        if(!cur)             {                res.push_back(lev);                lev.clear();                if(que.empty()) break;                que.push(NULL);            }            else            {                lev.push_back(cur->val);                if(cur->left) que.push(cur->left);                if(cur->right) que.push(cur->right);            }        }        return res;    }};