Question :
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.
Anwser 1 :
/**
* 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>> ret;
if(root == NULL) return ret;
vector<int> vec;
queue<TreeNode *> Q;
Q.push(root);
int count = 1;
while(!Q.empty()){
vec.clear();
int nextCount = 0; // cal next row count
for(int i = 0; i < count; i++){ // one row count
TreeNode *tmp = Q.front();
Q.pop();
vec.push_back(tmp->val); // save one row val
if(tmp->left){
Q.push(tmp->left);
nextCount++;
}
if(tmp->right){
Q.push(tmp->right);
nextCount++;
}
}
count = nextCount;
ret.push_back(vec);
}
return ret;
}
};
Anwser 2 :
/**
* 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>> ret;
if(root == NULL) return ret;
vector<int> vec;
queue<TreeNode *> Q;
queue<TreeNode *> Q2; // extra space
Q.push(root);
while(!Q.empty()){
TreeNode *tmp = Q.front();
Q.pop();
if(tmp != NULL){
vec.push_back(tmp->val);
if(tmp->left) Q2.push(tmp->left);
if(tmp->right) Q2.push(tmp->right);
}
if(Q.empty()){ // one row end
ret.push_back(vec);
vec.clear();
swap(Q, Q2);
}
}
}
};