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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<vector<int>> result;
if(!root) return result;
queue<TreeNode*> q1,q2;
q1.push(root);
TreeNode *cur;
vector<int> tmp;
while(!q1.empty()){
tmp.clear();
while(!q1.empty()){
cur = q1.front();
q1.pop();
tmp.push_back(cur -> val);
if(cur -> left) q2.push(cur -> left);
if(cur -> right) q2.push(cur -> right);
}
result.push_back(tmp);
swap(q1, q2);
}
return result;
}
};