feat(curriculum): restore seed + solution to Chinese (#40683)

* feat(tools): add seed/solution restore script

* chore(curriculum): remove empty sections' markers

* chore(curriculum): add seed + solution to Chinese

* chore: remove old formatter

* fix: update getChallenges

parse translated challenges separately, without reference to the source

* chore(curriculum): add dashedName to English

* chore(curriculum): add dashedName to Chinese

* refactor: remove unused challenge property 'name'

* fix: relax dashedName requirement

* fix: stray tag

Remove stray `pre` tag from challenge file.

Signed-off-by: nhcarrigan <nhcarrigan@gmail.com>

Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>

curriculum/challenges/chinese/10-coding-interview-prep/data-structures/invert-a-binary-tree.md

@@ -3,6 +3,7 @@ id: 587d8259367417b2b2512c83
 title: 反转二叉树
 challengeType: 1
 videoUrl: ''
+dashedName: invert-a-binary-tree
 ---
 
 # --description--
@@ -88,5 +89,104 @@ assert(
 );
 ```
 
+# --seed--
+
+## --after-user-code--
+
+```js
+BinarySearchTree.prototype = Object.assign(
+  BinarySearchTree.prototype,
+  {
+    add: function(value) {
+      function searchTree(node) {
+        if (value < node.value) {
+          if (node.left == null) {
+            node.left = new Node(value);
+            return;
+          } else if (node.left != null) {
+            return searchTree(node.left)
+          };
+        } else if (value > node.value) {
+          if (node.right == null) {
+            node.right = new Node(value);
+            return;
+          } else if (node.right != null) {
+            return searchTree(node.right);
+          };
+        } else {
+          return null;
+        };
+      }
+
+      var node = this.root;
+      if (node == null) {
+        this.root = new Node(value);
+        return;
+      } else {
+        return searchTree(node);
+      };
+    },
+    inorder: function() {
+      if (this.root == null) {
+        return null;
+      } else {
+        var result = new Array();
+        function traverseInOrder(node) {
+          if (node.left != null) {
+            traverseInOrder(node.left);
+          };
+          result.push(node.value);
+          if (node.right != null) {
+            traverseInOrder(node.right);
+          };
+        }
+        traverseInOrder(this.root);
+        return result;
+      };
+    }
+  }
+);
+```
+
+## --seed-contents--
+
+```js
+var displayTree = (tree) => console.log(JSON.stringify(tree, null, 2));
+function Node(value) {
+  this.value = value;
+  this.left = null;
+  this.right = null;
+}
+function BinarySearchTree() {
+  this.root = null;
+  // Only change code below this line
+  
+  // Only change code above this line
+}
+```
+
 # --solutions--
 
+```js
+var displayTree = (tree) => console.log(JSON.stringify(tree, null, 2));
+function Node(value) {
+  this.value = value;
+  this.left = null;
+  this.right = null;
+}
+function BinarySearchTree() {
+  this.root = null;
+  // Only change code below this line
+  this.invert = function(node = this.root) {
+    if (node) {
+      const temp = node.left;
+      node.left = node.right;
+      node.right = temp;
+      this.invert(node.left);
+      this.invert(node.right);
+    }
+    return node;
+  }
+    // Only change code above this line
+}
+```