标签：algorithm leetcode 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.
/**
* 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<int> vectorTemp;
vector<vector<int> > vectorTotal;
if(root==NULL)return vectorTotal;
queue<TreeNode*> Q;
TreeNode* p=root;
Q.push(p);
int counter=1;
while(!Q.empty())
{
p=Q.front();
Q.pop();
counter--;
vectorTemp.push_back(p->val);
if(p->left!=NULL)Q.push(p->left);
if(p->right!=NULL)Q.push(p->right);
if(counter==0)
{
//vector<int> vectorTemp;
vectorTotal.push_back(vectorTemp);
vectorTemp.clear();
counter=Q.size();
}
}
//vectorTotal.push_back(vectorTemp);
return vectorTotal;
}
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标签：algorithm leetcode traversal
原文地址：http://blog.csdn.net/plunger2011/article/details/25956357
