PS:这是我真正意义上一次AC的第一道题,以前出现过的虽然程序没问题,但是可能有手误打错出现编译错误,然后轻松修改提交。这次连手抖都没有了:)
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,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
使用了一个队列,代码如下:
/**
* 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:
vector<vector<int>> levelOrder(TreeNode* root) {
vector<vector<int>> res;
if(root == NULL)
return res;
queue<TreeNode*> que;
que.push(root);
while(!que.empty()){
vector<int> vec;
int size = que.size();
for(int i=0; i<size; ++i){
TreeNode* t = que.front();
que.pop();
vec.push_back(t->val);
if(t->left != NULL)
que.push(t->left);
if(t->right != NULL)
que.push(t->right);
}
res.push_back(vec);
}
return res;
}
};