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.
/**
* 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>> res;
if(root == NULL)
return res;
int save;
queue<TreeNode*> use;
use.push(root);
save = 1;
while(save!=0)
{
int count = save;
save = 0;
vector<int> t;
while(count > 0 && !use.empty())
{
TreeNode* temp = use.front();
use.pop();
count--;
t.push_back(temp->val);
if(temp->left!=NULL)
{
use.push(temp->left);
save++;
}
if(temp->right!=NULL)
{
use.push(temp->right);
save++;
}
}
res.push_back(t);
}
return res;
}
};LeetCode--Binary Tree Level Order Traversal
原文地址:http://blog.csdn.net/shaya118/article/details/42706477