mirror of
https://gitee.com/TheAlgorithms/LeetCodeAnimation.git
synced 2024-12-06 15:19:44 +08:00
99 lines
2.1 KiB
Markdown
99 lines
2.1 KiB
Markdown
# LeetCode 第 102 号问题：二叉树的层序遍历
|
||
|
||
> 本文首发于公众号「五分钟学算法」，是[图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>)系列文章之一。
|
||
>
|
||
> 个人网站：[https://www.cxyxiaowu.com](https://www.cxyxiaowu.com)
|
||
|
||
题目来源于 LeetCode 上第 102 号问题：二叉树的层序遍历。题目难度为 Medium，目前通过率为 55.8% 。
|
||
|
||
### 题目描述
|
||
|
||
给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
|
||
|
||
例如:
|
||
给定二叉树: `[3,9,20,null,null,15,7]`,
|
||
|
||
```
|
||
3
|
||
/ \
|
||
9 20
|
||
/ \
|
||
15 7
|
||
```
|
||
|
||
返回其层次遍历结果：
|
||
|
||
```
|
||
[
|
||
[3],
|
||
[9,20],
|
||
[15,7]
|
||
]
|
||
```
|
||
|
||
### 题目解析
|
||
|
||
该问题需要用到**队列**
|
||
|
||
- 建立一个queue
|
||
- 先把根节点放进去，这时候找根节点的左右两个子节点
|
||
- 去掉根节点，此时queue里的元素就是下一层的所有节点
|
||
- 用for循环遍历，将结果存到一个一维向量里
|
||
- 遍历完之后再把这个一维向量存到二维向量里
|
||
- 以此类推，可以完成层序遍历
|
||
|
||
|
||
|
||
|
||
|
||
### 动画描述
|
||
|
||
|
||
|
||
|
||
|
||
### 代码实现
|
||
|
||
```
|
||
/// BFS
|
||
/// Time Complexity: O(n), where n is the number of nodes in the tree
|
||
/// Space Complexity: O(n)
|
||
class Solution {
|
||
public:
|
||
vector<vector<int>> levelOrder(TreeNode* root) {
|
||
|
||
vector<vector<int>> res;
|
||
if(root == NULL)
|
||
return res;
|
||
|
||
queue<pair<TreeNode*,int>> q;
|
||
q.push(make_pair(root, 0));
|
||
|
||
while(!q.empty()){
|
||
|
||
TreeNode* node = q.front().first;
|
||
int level = q.front().second;
|
||
q.pop();
|
||
|
||
if(level == res.size())
|
||
res.push_back(vector<int>());
|
||
assert( level < res.size() );
|
||
|
||
res[level].push_back(node->val);
|
||
if(node->left)
|
||
q.push(make_pair(node->left, level + 1 ));
|
||
if(node->right)
|
||
q.push(make_pair(node->right, level + 1 ));
|
||
}
|
||
|
||
return res;
|
||
}
|
||
};
|
||
|
||
```
|
||
|
||
|
||
|
||
|
||
|
||
 |