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97 lines
2.1 KiB
Java
97 lines
2.1 KiB
Java
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# LeetCode 第 102 号问题：二叉树的层序遍历
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> 本文首发于公众号「图解面试算法」，是 [图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>) 系列文章之一。
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>
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> 同步博客：https://www.algomooc.com
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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% 。
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### 题目描述
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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）。
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例如:
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给定二叉树: `[3,9,20,null,null,15,7]`,
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```
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3
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/ \
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9 20
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/ \
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15 7
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```
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返回其层次遍历结果：
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```
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[
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[3],
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[9,20],
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[15,7]
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]
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```
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### 题目解析
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该问题需要用到**队列**
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- 建立一个queue
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- 先把根节点放进去，这时候找根节点的左右两个子节点
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- 去掉根节点，此时queue里的元素就是下一层的所有节点
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- 用for循环遍历，将结果存到一个一维向量里
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- 遍历完之后再把这个一维向量存到二维向量里
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- 以此类推，可以完成层序遍历
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### 动画描述
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### 代码实现
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```
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/// BFS
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/// Time Complexity: O(n), where n is the number of nodes in the tree
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/// Space Complexity: O(n)
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class Solution {
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public:
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vector<vector<int>> levelOrder(TreeNode* root) {
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vector<vector<int>> res;
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if(root == NULL)
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return res;
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queue<pair<TreeNode*,int>> q;
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q.push(make_pair(root, 0));
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while(!q.empty()){
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TreeNode* node = q.front().first;
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int level = q.front().second;
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q.pop();
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if(level == res.size())
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res.push_back(vector<int>());
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assert( level < res.size() );
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res[level].push_back(node->val);
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if(node->left)
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q.push(make_pair(node->left, level + 1 ));
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if(node->right)
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q.push(make_pair(node->right, level + 1 ));
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}
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return res;
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}
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};
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```
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