102.LeetCode Binary Tree Level Order Traversal(easy)[二叉树层次遍历 广度搜索 队列]

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]

]
解题思路：要求按照层次来遍历二叉树，可以采用队列广度搜索来完成，也可以采用下面的方法定义一个层次来完成。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void getResult(vector<vector<int> >&result,TreeNode* root,int l)
{
if(l==result.size())
{
vector<int> temp;
result.push_back(temp);
}
result[l].push_back(root->val);
if(root->left != NULL)
getResult(result,root->left,l+1);
if(root->right != NULL)
getResult(result,root->right,l+1);
}
vector<vector<int>> levelOrder(TreeNode* root) {
//用一个level来规定层数，这样可以保证同一层的在一个vector里面
vector<vector<int> > result;
if(root == NULL)
return result;
getResult(result,root,0);
return result;
}
};

<pre style="margin-top: 0px; margin-bottom: 10px; overflow: auto; font-family: Menlo, Monaco, Consolas, 'Courier New', monospace; font-size: 13px; padding: 9.5px; line-height: 1.42857; color: rgb(51, 51, 51); word-break: break-all; word-wrap: break-word; border: 1px solid rgb(204, 204, 204); background-color: rgb(245, 245, 245);"><pre code_snippet_id="1628381" snippet_file_name="blog_20160721_3_3617248" name="code" class="cpp">//用队列来实现，但是没有按照层次分组
vector<vector<int>> levelOrder(TreeNode* root) {
queue<TreeNode*> s;
vector<vector<int> > result;
if(root == NULL) return result;
s.push(root);
int i = 0;
vector<int> temp;
result.push_back(temp);
while(!s.empty())
{   //cout<<s.size()<<endl;
TreeNode *bt = NULL;
bt = s.front();
cout<<bt->val<<endl;
if(bt != NULL)
{
cout<<"in"<<endl;
result[1].push_back(bt->val);
s.pop();
s.push(bt->left);
s.push(bt->right);
}
// cout<<"xf"<<s.size()<<endl;
}
return result;
}