feat: add 2415

README.md

@@ -133,6 +133,7 @@
 - [2096. Step-By-Step Directions From a Binary Tree Node to Another(medium)](./book_sources/medium/2096.md)
 - [2191. Sort the Jumbled Numbers(medium)](./book_sources/medium/2191.md)
 - [2196. Create Binary Tree From Descriptions(medium)](./book_sources/medium/2196.md)
+- [2415. Reverse Odd Levels of Binary Tree(medium)](./book_sources/medium/2415.md)
 - [2418. Sort the People(easy)](./book_sources/easy/2418.md)
 - [2751. Robot Collisions(hard)](./book_sources/hard/2751.md)
 

book_sources/README.md

@@ -131,6 +131,7 @@
 - [2096. Step-By-Step Directions From a Binary Tree Node to Another(medium)](./medium/2096.md)
 - [2191. Sort the Jumbled Numbers(medium)](./medium/2191.md)
 - [2196. Create Binary Tree From Descriptions(medium)](./medium/2196.md)
+- [2415. Reverse Odd Levels of Binary Tree(medium)](./medium/2415.md)
 - [2418. Sort the People(easy)](./easy/2418.md)
 - [2751. Robot Collisions(hard)](./hard/2751.md)
 

book_sources/SUMMARY.md

@@ -120,6 +120,7 @@
   - [2096. Step-By-Step Directions From a Binary Tree Node to Another(medium)](./medium/2096.md)
   - [2191. Sort the Jumbled Numbers(medium)](./medium/2191.md)
   - [2196. Create Binary Tree From Descriptions(medium)](./medium/2196.md)
+  - [2415. Reverse Odd Levels of Binary Tree(medium)](./medium/2415.md)
 - [Hard](hard/README.md)
   - [32. Longest Valid Parentheses(hard)](./hard/32.md)
   - [51. N-Queens(hard)](./hard/51.md)

book_sources/medium/2415.md

@@ -0,0 +1,57 @@
+# 2415. Reverse Odd Levels of Binary Tree
+
+> [Leetcode link](https://leetcode.com/problems/reverse-odd-levels-of-binary-tree)
+
+## 题目简介
+
+题目给我们一个完美二叉树（所有父节点都有两个子节点且所有叶子节点都在同一层），要求我们将奇数层的所有叶子节点翻转（简单理解成 Array.reverse() 就好
+
+## 解题思路
+
+其实翻转的本质可以看成，有两个指针分别指向需要交换的层的最左边与最右边的节点，然后将其交换后把指针各自往中间走一步，直到指针相遇
+
+那么我们在树中要怎么样找到当前层最左跟最右的节点呢？答案是 dfs
+
+我们只需要在深度遍历的时候，同时遍历两个节点就好了，所以 dfs 的调用应该是这样的：
+
+`dfs(node.left, node.right, level)`
+
+那么我们接下来要怎么让两个 “指针” 同时往中间走呢？答案是反过来遍历：
+
+`dfs(node.right, node.left, level)`
+
+基于以上，我们可以看代码了
+
+### Javascript
+
+```js
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val, left, right) {
+ *     this.val = (val===undefined ? 0 : val)
+ *     this.left = (left===undefined ? null : left)
+ *     this.right = (right===undefined ? null : right)
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {TreeNode}
+ */
+var reverseOddLevels = function(root) {
+    const dfs = (left, right, level) => {
+        if(left === null || right === null) {
+            return
+        }
+        if(level % 2 === 1) {
+            [left.val, right.val] = [right.val, left.val]
+        }
+        dfs(left.left, right.right, level+1)
+        dfs(left.right, right.left, level+1)
+    }
+
+    dfs(root.left, root.right, 1)
+    return root
+
+};
+```
+