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feat: add solutions to leetcode problem: No.0102
See https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
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‎README.md

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@@ -84,6 +84,7 @@
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### 二叉树
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1. [对称二叉树](/solution/0100-0199/0101.Symmetric%20Tree/README.md)
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1. [二叉树的层次遍历](/solution/0100-0199/0102.Binary%20Tree%20Level%20Order%20Traversal/README.md)
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1. [二叉树的最大深度](/solution/0100-0199/0104.Maximum%20Depth%20of%20Binary%20Tree/README.md)
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1. [二叉树的最近公共祖先](/solution/0200-0299/0235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree/README.md)
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1. [二叉搜索树的最近公共祖先](/solution/0200-0299/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/README.md)

‎README_EN.md

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@@ -82,6 +82,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
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### Binary Tree
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1. [Symmetric Tree](/solution/0100-0199/0101.Symmetric%20Tree/README_EN.md)
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1. [Binary Tree Level Order Traversal](/solution/0100-0199/0102.Binary%20Tree%20Level%20Order%20Traversal/README_EN.md)
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1. [Maximum Depth of Binary Tree](/solution/0100-0199/0104.Maximum%20Depth%20of%20Binary%20Tree/README_EN.md)
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1. [Lowest Common Ancestor of a Binary Tree](/solution/0200-0299/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/README_EN.md)
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1. [Lowest Common Ancestor of a Binary Search Tree](/solution/0200-0299/0235.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Search%20Tree/README_EN.md)

‎solution/0100-0199/0102.Binary Tree Level Order Traversal/README.md

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@@ -30,22 +30,111 @@
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<!-- 这里可写通用的实现逻辑 -->
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队列实现。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def levelOrder(self, root: TreeNode) -> List[List[int]]:
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if root is None:
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return []
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res = []
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q = []
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q.append(root)
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while q:
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size = len(q)
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t = []
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for _ in range(size):
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node = q.pop(0)
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if node.left is not None:
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q.append(node.left)
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if node.right is not None:
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q.append(node.right)
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t.append(node.val)
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res.append(t)
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return res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public List<List<Integer>> levelOrder(TreeNode root) {
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if (root == null) return Collections.emptyList();
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Deque<TreeNode> q = new ArrayDeque<>();
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q.offer(root);
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List<List<Integer>> res = new ArrayList<>();
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while (!q.isEmpty()) {
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int size = q.size();
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List<Integer> t = new ArrayList<>();
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while (size-- > 0) {
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TreeNode node = q.poll();
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t.add(node.val);
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if (node.left != null) q.offer(node.left);
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if (node.right != null) q.offer(node.right);
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}
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res.add(t);
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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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### **C++**
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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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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if (!root) return {};
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vector<vector<int>> res;
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queue<TreeNode*> q{{root}};
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while (!q.empty()) {
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vector<int> oneLevel;
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for (int i = q.size(); i > 0; --i) {
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TreeNode* t = q.front();
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q.pop();
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oneLevel.push_back(t->val);
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if (t->left) q.push(t->left);
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if (t->right) q.push(t->right);
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}
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res.push_back(oneLevel);
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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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### **...**

‎solution/0100-0199/0102.Binary Tree Level Order Traversal/README_EN.md

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@@ -55,13 +55,100 @@ return its level order traversal as:<br />
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### **Python3**
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def levelOrder(self, root: TreeNode) -> List[List[int]]:
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if root is None:
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return []
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res = []
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q = []
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q.append(root)
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while q:
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size = len(q)
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t = []
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for _ in range(size):
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node = q.pop(0)
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if node.left is not None:
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q.append(node.left)
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if node.right is not None:
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q.append(node.right)
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t.append(node.val)
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res.append(t)
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return res
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```
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### **Java**
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public List<List<Integer>> levelOrder(TreeNode root) {
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if (root == null) return Collections.emptyList();
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Deque<TreeNode> q = new ArrayDeque<>();
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q.offer(root);
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List<List<Integer>> res = new ArrayList<>();
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while (!q.isEmpty()) {
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int size = q.size();
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List<Integer> t = new ArrayList<>();
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while (size-- > 0) {
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TreeNode node = q.poll();
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t.add(node.val);
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if (node.left != null) q.offer(node.left);
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if (node.right != null) q.offer(node.right);
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}
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res.add(t);
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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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### **C++**
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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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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if (!root) return {};
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vector<vector<int>> res;
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queue<TreeNode*> q{{root}};
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while (!q.empty()) {
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vector<int> oneLevel;
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for (int i = q.size(); i > 0; --i) {
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TreeNode* t = q.front();
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q.pop();
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oneLevel.push_back(t->val);
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if (t->left) q.push(t->left);
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if (t->right) q.push(t->right);
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}
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res.push_back(oneLevel);
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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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### **...**

‎solution/0100-0199/0102.Binary Tree Level Order Traversal/Solution.java

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@@ -1,28 +1,29 @@
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public List<List<Integer>> levelOrder(TreeNode root) {
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if (root == null) return Collections.emptyList();
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Deque<TreeNode> q = new ArrayDeque<>();
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q.offer(root);
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List<List<Integer>> res = new ArrayList<>();
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if (root == null) {
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return res;
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}
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Queue<TreeNode> queue = new LinkedList<>();
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queue.offer(root);
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while (!queue.isEmpty()) {
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int size = queue.size();
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List<Integer> list = new ArrayList<>();
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while ((size--) > 0) {
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TreeNode node = queue.poll();
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list.add(node.val);
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if (node.left != null) {
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queue.offer(node.left);
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}
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if (node.right != null) {
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queue.offer(node.right);
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}
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while (!q.isEmpty()) {
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int size = q.size();
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List<Integer> t = new ArrayList<>();
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while (size-- > 0) {
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TreeNode node = q.poll();
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t.add(node.val);
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if (node.left != null) q.offer(node.left);
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if (node.right != null) q.offer(node.right);
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}
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res.add(list);
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res.add(t);
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}
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return res;
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}
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}

‎solution/0100-0199/0102.Binary Tree Level Order Traversal/Solution.py

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# self.right = None
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class Solution:
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def level(self, nodes):
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"""
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:type nodes: List[TreeNode]
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:rtype: List[List[int]]
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"""
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Ret = [[]]
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Next_Level = []
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for each in nodes:
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Ret[0].append(each.val)
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if(each.left != None):
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Next_Level.append(each.left)
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if(each.right != None):
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Next_Level.append(each.right)
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if Next_Level :
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Ret.extend(self.level(Next_Level))
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return Ret
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def levelOrder(self, root):
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"""
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:type root: TreeNode
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:rtype: List[List[int]]
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"""
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if(root != None):
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return self.level([root])
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else:
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def levelOrder(self, root: TreeNode) -> List[List[int]]:
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if root is None:
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return []
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res = []
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q = []
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q.append(root)
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while q:
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size = len(q)
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t = []
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for _ in range(size):
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node = q.pop(0)
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if node.left is not None:
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q.append(node.left)
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if node.right is not None:
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q.append(node.right)
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t.append(node.val)
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res.append(t)
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return res

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